Calculus Examples

$f (x) = x^{2} e^{4 x}$

Step 1

Find the second derivative.

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Find the first derivative.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = e^{4 x}$ .

$f' (x) = x^{2} \frac{d}{d x} (e^{4 x}) + e^{4 x} \frac{d}{d x} (x^{2})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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To apply the Chain Rule, set $u$ as $4 x$ .

$f' (x) = x^{2} (\frac{d}{d u} (e^{u}) \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x^{2})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$f' (x) = x^{2} (e^{u} \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x^{2})$

Replace all occurrences of $u$ with $4 x$ .

$f' (x) = x^{2} (e^{4 x} \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x^{2})$

Differentiate.

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Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f' (x) = x^{2} (e^{4 x} (4 \frac{d}{d x} (x))) + e^{4 x} \frac{d}{d x} (x^{2})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = x^{2} (e^{4 x} (4 \cdot 1)) + e^{4 x} \frac{d}{d x} (x^{2})$

Simplify the expression.

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Multiply $4$ by $1$ .

$f' (x) = x^{2} (e^{4 x} \cdot 4) + e^{4 x} \frac{d}{d x} (x^{2})$

Move $4$ to the left of $e^{4 x}$ .

$f' (x) = x^{2} (4 \cdot e^{4 x}) + e^{4 x} \frac{d}{d x} (x^{2})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = x^{2} (4 e^{4 x}) + e^{4 x} (2 x)$

Simplify.

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Reorder terms.

$f' (x) = 4 e^{4 x} x^{2} + 2 e^{4 x} x$

Reorder factors in $4 e^{4 x} x^{2} + 2 e^{4 x} x$ .

$f' (x) = 4 x^{2} e^{4 x} + 2 x e^{4 x}$

Find the second derivative.

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By the Sum Rule, the derivative of $4 x^{2} e^{4 x} + 2 x e^{4 x}$ with respect to $x$ is $\frac{d}{d x} [4 x^{2} e^{4 x}] + \frac{d}{d x} [2 x e^{4 x}]$ .

$f'' (x) = \frac{d}{d x} (4 x^{2} e^{4 x}) + \frac{d}{d x} (2 x e^{4 x})$

Evaluate $\frac{d}{d x} [4 x^{2} e^{4 x}]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2} e^{4 x}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2} e^{4 x}]$ .

$f'' (x) = 4 \frac{d}{d x} (x^{2} e^{4 x}) + \frac{d}{d x} (2 x e^{4 x})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = e^{4 x}$ .

$f'' (x) = 4 (x^{2} \frac{d}{d x} (e^{4 x}) + e^{4 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{4 x})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$f'' (x) = 4 (x^{2} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{4 x})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = 4 (x^{2} (e^{u_{1}} \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{4 x})$

Replace all occurrences of $u_{1}$ with $4 x$ .

$f'' (x) = 4 (x^{2} (e^{4 x} \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{4 x})$

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f'' (x) = 4 (x^{2} (e^{4 x} (4 \frac{d}{d x} (x))) + e^{4 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{4 x})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 4 (x^{2} (e^{4 x} (4 \cdot 1)) + e^{4 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{4 x})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 4 (x^{2} (e^{4 x} (4 \cdot 1)) + e^{4 x} (2 x)) + \frac{d}{d x} (2 x e^{4 x})$

Multiply $4$ by $1$ .

$f'' (x) = 4 (x^{2} (e^{4 x} \cdot 4) + e^{4 x} (2 x)) + \frac{d}{d x} (2 x e^{4 x})$

Move $4$ to the left of $e^{4 x}$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + \frac{d}{d x} (2 x e^{4 x})$

Evaluate $\frac{d}{d x} [2 x e^{4 x}]$ .

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Since $2$ is constant with respect to $x$ , the derivative of $2 x e^{4 x}$ with respect to $x$ is $2 \frac{d}{d x} [x e^{4 x}]$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 \frac{d}{d x} (x e^{4 x})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{4 x}$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x \frac{d}{d x} (e^{4 x}) + e^{4 x} \frac{d}{d x} (x))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (e^{u_{2}} \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x))$

Replace all occurrences of $u_{2}$ with $4 x$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (e^{4 x} \frac{d}{d x} (4 x)) + e^{4 x} \frac{d}{d x} (x))$

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (e^{4 x} (4 \frac{d}{d x} (x))) + e^{4 x} \frac{d}{d x} (x))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (e^{4 x} (4 \cdot 1)) + e^{4 x} \frac{d}{d x} (x))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (e^{4 x} (4 \cdot 1)) + e^{4 x} \cdot 1)$

Multiply $4$ by $1$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (e^{4 x} \cdot 4) + e^{4 x} \cdot 1)$

Move $4$ to the left of $e^{4 x}$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (4 \cdot e^{4 x}) + e^{4 x} \cdot 1)$

Multiply $e^{4 x}$ by $1$ .

$f'' (x) = 4 (x^{2} (4 e^{4 x}) + e^{4 x} (2 x)) + 2 (x (4 e^{4 x}) + e^{4 x})$

Simplify.

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Apply the distributive property.

$f'' (x) = 4 (x^{2} (4 e^{4 x})) + 4 (e^{4 x} (2 x)) + 2 (x (4 e^{4 x}) + e^{4 x})$

Apply the distributive property.

$f'' (x) = 4 (x^{2} (4 e^{4 x})) + 4 (e^{4 x} (2 x)) + 2 (x (4 e^{4 x})) + 2 e^{4 x}$

Combine terms.

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Multiply $4$ by $4$ .

$f'' (x) = 16 (x^{2} (e^{4 x})) + 4 (e^{4 x} (2 x)) + 2 (x (4 e^{4 x})) + 2 e^{4 x}$

Multiply $2$ by $4$ .

$f'' (x) = 16 x^{2} e^{4 x} + 8 (e^{4 x} (x)) + 2 (x (4 e^{4 x})) + 2 e^{4 x}$

Multiply $4$ by $2$ .

$f'' (x) = 16 x^{2} e^{4 x} + 8 e^{4 x} x + 8 (x (e^{4 x})) + 2 e^{4 x}$

Add $8 e^{4 x} x$ and $8 x e^{4 x}$ .

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Move $e^{4 x}$ .

$f'' (x) = 16 x^{2} e^{4 x} + 8 x e^{4 x} + 8 x e^{4 x} + 2 e^{4 x}$

Add $8 x e^{4 x}$ and $8 x e^{4 x}$ .

$f'' (x) = 16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x}$

Reorder terms.

$f'' (x) = 16 e^{4 x} x^{2} + 16 e^{4 x} x + 2 e^{4 x}$

Reorder factors in $16 e^{4 x} x^{2} + 16 e^{4 x} x + 2 e^{4 x}$ .

$f'' (x) = 16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x}$

The second derivative of $f (x)$ with respect to $x$ is $16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x}$ .

$16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x}$

Step 2

Set the second derivative equal to $0$ then solve the equation $16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x} = 0$ .

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Set the second derivative equal to $0$ .

$16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x} = 0$

Factor $2 e^{4 x}$ out of $16 x^{2} e^{4 x} + 16 x e^{4 x} + 2 e^{4 x}$ .

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Factor $2 e^{4 x}$ out of $16 x^{2} e^{4 x}$ .

$2 e^{4 x} (8 x^{2}) + 16 x e^{4 x} + 2 e^{4 x} = 0$

Factor $2 e^{4 x}$ out of $16 x e^{4 x}$ .

$2 e^{4 x} (8 x^{2}) + 2 e^{4 x} (8 x) + 2 e^{4 x} = 0$

Factor $2 e^{4 x}$ out of $2 e^{4 x}$ .

$2 e^{4 x} (8 x^{2}) + 2 e^{4 x} (8 x) + 2 e^{4 x} (1) = 0$

Factor $2 e^{4 x}$ out of $2 e^{4 x} (8 x^{2}) + 2 e^{4 x} (8 x)$ .

$2 e^{4 x} (8 x^{2} + 8 x) + 2 e^{4 x} (1) = 0$

Factor $2 e^{4 x}$ out of $2 e^{4 x} (8 x^{2} + 8 x) + 2 e^{4 x} (1)$ .

$2 e^{4 x} (8 x^{2} + 8 x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{4 x} = 0$

$8 x^{2} + 8 x + 1 = 0$

Set $e^{4 x}$ equal to $0$ and solve for $x$ .

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Set $e^{4 x}$ equal to $0$ .

$e^{4 x} = 0$

Solve $e^{4 x} = 0$ for $x$ .

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Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{4 x}) = \ln (0)$

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

There is no solution for $e^{4 x} = 0$

No solution

Set $8 x^{2} + 8 x + 1$ equal to $0$ and solve for $x$ .

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Set $8 x^{2} + 8 x + 1$ equal to $0$ .

$8 x^{2} + 8 x + 1 = 0$

Solve $8 x^{2} + 8 x + 1 = 0$ for $x$ .

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Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 8$ , $b = 8$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (8 \cdot 1)}}{2 \cdot 8}$

Simplify.

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Simplify the numerator.

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Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 8 \cdot 1}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot 1$ .

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Multiply $- 4$ by $8$ .

$x = \frac{- 8 \pm \sqrt{64 - 32 \cdot 1}}{2 \cdot 8}$

Multiply $- 32$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 32}}{2 \cdot 8}$

Subtract $32$ from $64$ .

$x = \frac{- 8 \pm \sqrt{32}}{2 \cdot 8}$

Rewrite $32$ as $4^{2} \cdot 2$ .

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Factor $16$ out of $32$ .

$x = \frac{- 8 \pm \sqrt{16 (2)}}{2 \cdot 8}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 8 \pm \sqrt{4^{2} \cdot 2}}{2 \cdot 8}$

Pull terms out from under the radical.

$x = \frac{- 8 \pm 4 \sqrt{2}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 8 \pm 4 \sqrt{2}}{16}$

Simplify $\frac{- 8 \pm 4 \sqrt{2}}{16}$ .

$x = \frac{- 2 \pm \sqrt{2}}{4}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 8 \cdot 1}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot 1$ .

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Multiply $- 4$ by $8$ .

$x = \frac{- 8 \pm \sqrt{64 - 32 \cdot 1}}{2 \cdot 8}$

Multiply $- 32$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 32}}{2 \cdot 8}$

Subtract $32$ from $64$ .

$x = \frac{- 8 \pm \sqrt{32}}{2 \cdot 8}$

Rewrite $32$ as $4^{2} \cdot 2$ .

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Factor $16$ out of $32$ .

$x = \frac{- 8 \pm \sqrt{16 (2)}}{2 \cdot 8}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 8 \pm \sqrt{4^{2} \cdot 2}}{2 \cdot 8}$

Pull terms out from under the radical.

$x = \frac{- 8 \pm 4 \sqrt{2}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 8 \pm 4 \sqrt{2}}{16}$

Simplify $\frac{- 8 \pm 4 \sqrt{2}}{16}$ .

$x = \frac{- 2 \pm \sqrt{2}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{- 2 + \sqrt{2}}{4}$

Rewrite $- 2$ as $- 1 (2)$ .

$x = \frac{- 1 \cdot 2 + \sqrt{2}}{4}$

Factor $- 1$ out of $\sqrt{2}$ .

$x = \frac{- 1 \cdot 2 - 1 (- \sqrt{2})}{4}$

Factor $- 1$ out of $- 1 (2) - 1 (- \sqrt{2})$ .

$x = \frac{- 1 (2 - \sqrt{2})}{4}$

Move the negative in front of the fraction.

$x = - \frac{2 - \sqrt{2}}{4}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 8 \cdot 1}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot 1$ .

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Multiply $- 4$ by $8$ .

$x = \frac{- 8 \pm \sqrt{64 - 32 \cdot 1}}{2 \cdot 8}$

Multiply $- 32$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 32}}{2 \cdot 8}$

Subtract $32$ from $64$ .

$x = \frac{- 8 \pm \sqrt{32}}{2 \cdot 8}$

Rewrite $32$ as $4^{2} \cdot 2$ .

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Factor $16$ out of $32$ .

$x = \frac{- 8 \pm \sqrt{16 (2)}}{2 \cdot 8}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 8 \pm \sqrt{4^{2} \cdot 2}}{2 \cdot 8}$

Pull terms out from under the radical.

$x = \frac{- 8 \pm 4 \sqrt{2}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 8 \pm 4 \sqrt{2}}{16}$

Simplify $\frac{- 8 \pm 4 \sqrt{2}}{16}$ .

$x = \frac{- 2 \pm \sqrt{2}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{- 2 - \sqrt{2}}{4}$

Rewrite $- 2$ as $- 1 (2)$ .

$x = \frac{- 1 \cdot 2 - \sqrt{2}}{4}$

Factor $- 1$ out of $- \sqrt{2}$ .

$x = \frac{- 1 \cdot 2 - (\sqrt{2})}{4}$

Factor $- 1$ out of $- 1 (2) - (\sqrt{2})$ .

$x = \frac{- 1 (2 + \sqrt{2})}{4}$

Move the negative in front of the fraction.

$x = - \frac{2 + \sqrt{2}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{2 - \sqrt{2}}{4}, - \frac{2 + \sqrt{2}}{4}$

The final solution is all the values that make $2 e^{4 x} (8 x^{2} + 8 x + 1) = 0$ true.

$x = - \frac{2 - \sqrt{2}}{4}, - \frac{2 + \sqrt{2}}{4}$

Step 3

Find the points where the second derivative is $0$ .

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Substitute $- 0.1464466$ in $f (x) = x^{2} e^{4 x}$ to find the value of $y$ .

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Replace the variable $x$ with $- 0.1464466$ in the expression.

$f (- 0.1464466) = {(- 0.1464466)}^{2} e^{4 (- 0.1464466)}$

Simplify the result.

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Raise $- 0.1464466$ to the power of $2$ .

$f (- 0.1464466) = 0.0214466 e^{4 (- 0.1464466)}$

Multiply $4$ by $- 0.1464466$ .

$f (- 0.1464466) = 0.0214466 e^{- 0.58578643}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f (- 0.1464466) = 0.0214466 (\frac{1}{e^{0.58578643}})$

Combine $0.0214466$ and $\frac{1}{e^{0.58578643}}$ .

$f (- 0.1464466) = \frac{0.0214466}{e^{0.58578643}}$

Replace $e$ with an approximation.

$f (- 0.1464466) = \frac{0.0214466}{{2.71828182}^{0.58578643}}$

Raise $2.71828182$ to the power of $0.58578643$ .

$f (- 0.1464466) = \frac{0.0214466}{1.79640318}$

Divide $0.0214466$ by $1.79640318$ .

$f (- 0.1464466) = 0.01193863$

The final answer is $0.01193863$ .

$0.01193863$

The point found by substituting $- 0.1464466$ in $f (x) = x^{2} e^{4 x}$ is $(- 0.1464466, 0.01193863)$ . This point can be an inflection point.

$(- 0.1464466, 0.01193863)$

Substitute $- 0.85355339$ in $f (x) = x^{2} e^{4 x}$ to find the value of $y$ .

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Replace the variable $x$ with $- 0.85355339$ in the expression.

$f (- 0.85355339) = {(- 0.85355339)}^{2} e^{4 (- 0.85355339)}$

Simplify the result.

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Raise $- 0.85355339$ to the power of $2$ .

$f (- 0.85355339) = 0.72855339 e^{4 (- 0.85355339)}$

Multiply $4$ by $- 0.85355339$ .

$f (- 0.85355339) = 0.72855339 e^{- 3.41421356}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f (- 0.85355339) = 0.72855339 (\frac{1}{e^{3.41421356}})$

Combine $0.72855339$ and $\frac{1}{e^{3.41421356}}$ .

$f (- 0.85355339) = \frac{0.72855339}{e^{3.41421356}}$

Replace $e$ with an approximation.

$f (- 0.85355339) = \frac{0.72855339}{{2.71828182}^{3.41421356}}$

Raise $2.71828182$ to the power of $3.41421356$ .

$f (- 0.85355339) = \frac{0.72855339}{30.39303779}$

Divide $0.72855339$ by $30.39303779$ .

$f (- 0.85355339) = 0.02397106$

The final answer is $0.02397106$ .

$0.02397106$

The point found by substituting $- 0.85355339$ in $f (x) = x^{2} e^{4 x}$ is $(- 0.85355339, 0.02397106)$ . This point can be an inflection point.

$(- 0.85355339, 0.02397106)$

Determine the points that could be inflection points.

$(- 0.1464466, 0.01193863), (- 0.85355339, 0.02397106)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 0.85355339) \cup (- 0.85355339, - 0.1464466) \cup (- 0.1464466, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 0.85355339)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.95355339$ in the expression.

$f'' (- 0.95355339) = 16 {(- 0.95355339)}^{2} e^{4 (- 0.95355339)} + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Simplify the result.

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Simplify each term.

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Raise $- 0.95355339$ to the power of $2$ .

$f'' (- 0.95355339) = 16 \cdot (0.90926406 e^{4 (- 0.95355339)}) + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Multiply $16$ by $0.90926406$ .

$f'' (- 0.95355339) = 14.54822509 e^{4 (- 0.95355339)} + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Multiply $4$ by $- 0.95355339$ .

$f'' (- 0.95355339) = 14.54822509 e^{- 3.81421356} + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- 0.95355339) = 14.54822509 (\frac{1}{e^{3.81421356}}) + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Combine $14.54822509$ and $\frac{1}{e^{3.81421356}}$ .

$f'' (- 0.95355339) = \frac{14.54822509}{e^{3.81421356}} + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Replace $e$ with an approximation.

$f'' (- 0.95355339) = \frac{14.54822509}{{2.71828182}^{3.81421356}} + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Raise $2.71828182$ to the power of $3.81421356$ .

$f'' (- 0.95355339) = \frac{14.54822509}{45.34108442} + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Divide $14.54822509$ by $45.34108442$ .

$f'' (- 0.95355339) = 0.32086186 + 16 (- 0.95355339) \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Multiply $16$ by $- 0.95355339$ .

$f'' (- 0.95355339) = 0.32086186 - 15.25685424 \cdot e^{4 (- 0.95355339)} + 2 e^{4 (- 0.95355339)}$

Multiply $4$ by $- 0.95355339$ .

$f'' (- 0.95355339) = 0.32086186 - 15.25685424 \cdot e^{- 3.81421356} + 2 e^{4 (- 0.95355339)}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- 0.95355339) = 0.32086186 - 15.25685424 \cdot \frac{1}{e^{3.81421356}} + 2 e^{4 (- 0.95355339)}$

Combine $- 15.25685424$ and $\frac{1}{e^{3.81421356}}$ .

$f'' (- 0.95355339) = 0.32086186 + \frac{- 15.25685424}{e^{3.81421356}} + 2 e^{4 (- 0.95355339)}$

Move the negative in front of the fraction.

$f'' (- 0.95355339) = 0.32086186 - \frac{15.25685424}{e^{3.81421356}} + 2 e^{4 (- 0.95355339)}$

Replace $e$ with an approximation.

$f'' (- 0.95355339) = 0.32086186 - \frac{15.25685424}{{2.71828182}^{3.81421356}} + 2 e^{4 (- 0.95355339)}$

Raise $2.71828182$ to the power of $3.81421356$ .

$f'' (- 0.95355339) = 0.32086186 - \frac{15.25685424}{45.34108442} + 2 e^{4 (- 0.95355339)}$

Divide $15.25685424$ by $45.34108442$ .

$f'' (- 0.95355339) = 0.32086186 - 1 \cdot 0.33649072 + 2 e^{4 (- 0.95355339)}$

Multiply $- 1$ by $0.33649072$ .

$f'' (- 0.95355339) = 0.32086186 - 0.33649072 + 2 e^{4 (- 0.95355339)}$

Multiply $4$ by $- 0.95355339$ .

$f'' (- 0.95355339) = 0.32086186 - 0.33649072 + 2 e^{- 3.81421356}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- 0.95355339) = 0.32086186 - 0.33649072 + 2 (\frac{1}{e^{3.81421356}})$

Combine $2$ and $\frac{1}{e^{3.81421356}}$ .

$f'' (- 0.95355339) = 0.32086186 - 0.33649072 + \frac{2}{e^{3.81421356}}$

Simplify by adding terms.

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Subtract $0.33649072$ from $0.32086186$ .

$f'' (- 0.95355339) = - 0.01562885 + \frac{2}{e^{3.81421356}}$

Add $- 0.01562885$ and $\frac{2}{e^{3.81421356}}$ .

$f'' (- 0.95355339) = 0.02848125$

The final answer is $0.02848125$ .

$0.02848125$

At $- 0.95355339$ , the second derivative is $0.02848125$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 0.85355339)$ .

Increasing on $(- \infty, - 0.85355339)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(- 0.85355339, - 0.1464466)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f'' (- \frac{1}{2}) = 16 {(- \frac{1}{2})}^{2} e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Simplify the result.

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Simplify each term.

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Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1}{2}$ .

$f'' (- \frac{1}{2}) = 16 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Apply the product rule to $\frac{1}{2}$ .

$f'' (- \frac{1}{2}) = 16 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}})) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Raise $- 1$ to the power of $2$ .

$f'' (- \frac{1}{2}) = 16 (1 (\frac{1^{2}}{2^{2}})) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

$f'' (- \frac{1}{2}) = 16 (\frac{1^{2}}{2^{2}}) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

One to any power is one.

$f'' (- \frac{1}{2}) = 16 (\frac{1}{2^{2}}) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Raise $2$ to the power of $2$ .

$f'' (- \frac{1}{2}) = 16 (\frac{1}{4}) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor of $4$ .

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Factor $4$ out of $16$ .

$f'' (- \frac{1}{2}) = 4 (4) (\frac{1}{4}) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor.

$f'' (- \frac{1}{2}) = 4 \cdot (4 (\frac{1}{4})) \cdot e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Rewrite the expression.

$f'' (- \frac{1}{2}) = 4 e^{4 (- \frac{1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f'' (- \frac{1}{2}) = 4 e^{4 (\frac{- 1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Factor $2$ out of $4$ .

$f'' (- \frac{1}{2}) = 4 e^{2 (2) (\frac{- 1}{2})} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor.

$f'' (- \frac{1}{2}) = 4 e^{2 \cdot (2 (\frac{- 1}{2}))} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Rewrite the expression.

$f'' (- \frac{1}{2}) = 4 e^{2 \cdot - 1} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Multiply $2$ by $- 1$ .

$f'' (- \frac{1}{2}) = 4 e^{- 2} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- \frac{1}{2}) = 4 (\frac{1}{e^{2}}) + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Combine $4$ and $\frac{1}{e^{2}}$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} + 16 (- \frac{1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} + 16 (\frac{- 1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Factor $2$ out of $16$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} + 2 (8) (\frac{- 1}{2}) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} + 2 \cdot (8 (\frac{- 1}{2})) \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Rewrite the expression.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} + 8 \cdot - 1 \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Multiply $8$ by $- 1$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot e^{4 (- \frac{1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot e^{4 (\frac{- 1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Factor $2$ out of $4$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot e^{2 (2) (\frac{- 1}{2})} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot e^{2 \cdot (2 (\frac{- 1}{2}))} + 2 e^{4 (- \frac{1}{2})}$

Rewrite the expression.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot e^{2 \cdot - 1} + 2 e^{4 (- \frac{1}{2})}$

Multiply $2$ by $- 1$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot e^{- 2} + 2 e^{4 (- \frac{1}{2})}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - 8 \cdot \frac{1}{e^{2}} + 2 e^{4 (- \frac{1}{2})}$

Combine $- 8$ and $\frac{1}{e^{2}}$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} + \frac{- 8}{e^{2}} + 2 e^{4 (- \frac{1}{2})}$

Move the negative in front of the fraction.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{4 (- \frac{1}{2})}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{4 (\frac{- 1}{2})}$

Factor $2$ out of $4$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{2 (2) (\frac{- 1}{2})}$

Cancel the common factor.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{2 \cdot (2 (\frac{- 1}{2}))}$

Rewrite the expression.

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{- 2}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 (\frac{1}{e^{2}})$

Combine $2$ and $\frac{1}{e^{2}}$ .

$f'' (- \frac{1}{2}) = \frac{4}{e^{2}} - \frac{8}{e^{2}} + \frac{2}{e^{2}}$

Combine fractions.

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Combine the numerators over the common denominator.

$f'' (- \frac{1}{2}) = \frac{4 - 8 + 2}{e^{2}}$

Simplify the expression.

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Subtract $8$ from $4$ .

$f'' (- \frac{1}{2}) = \frac{- 4 + 2}{e^{2}}$

Add $- 4$ and $2$ .

$f'' (- \frac{1}{2}) = \frac{- 2}{e^{2}}$

Move the negative in front of the fraction.

$f'' (- \frac{1}{2}) = - \frac{2}{e^{2}}$

The final answer is $- \frac{2}{e^{2}}$ .

$- \frac{2}{e^{2}}$

At $- \frac{1}{2}$ , the second derivative is $- 0.27067056$ . Since this is negative, the second derivative is decreasing on the interval $(- 0.85355339, - 0.1464466)$

Decreasing on $(- 0.85355339, - 0.1464466)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(- 0.1464466, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.0464466$ in the expression.

$f'' (- 0.0464466) = 16 {(- 0.0464466)}^{2} e^{4 (- 0.0464466)} + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Simplify the result.

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Simplify each term.

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Raise $- 0.0464466$ to the power of $2$ .

$f'' (- 0.0464466) = 16 \cdot (0.00215728 e^{4 (- 0.0464466)}) + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Multiply $16$ by $0.00215728$ .

$f'' (- 0.0464466) = 0.0345166 e^{4 (- 0.0464466)} + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Multiply $4$ by $- 0.0464466$ .

$f'' (- 0.0464466) = 0.0345166 e^{- 0.18578643} + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- 0.0464466) = 0.0345166 (\frac{1}{e^{0.18578643}}) + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Combine $0.0345166$ and $\frac{1}{e^{0.18578643}}$ .

$f'' (- 0.0464466) = \frac{0.0345166}{e^{0.18578643}} + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Replace $e$ with an approximation.

$f'' (- 0.0464466) = \frac{0.0345166}{{2.71828182}^{0.18578643}} + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Raise $2.71828182$ to the power of $0.18578643$ .

$f'' (- 0.0464466) = \frac{0.0345166}{1.20416506} + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Divide $0.0345166$ by $1.20416506$ .

$f'' (- 0.0464466) = 0.02866434 + 16 (- 0.0464466) \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Multiply $16$ by $- 0.0464466$ .

$f'' (- 0.0464466) = 0.02866434 - 0.74314575 \cdot e^{4 (- 0.0464466)} + 2 e^{4 (- 0.0464466)}$

Multiply $4$ by $- 0.0464466$ .

$f'' (- 0.0464466) = 0.02866434 - 0.74314575 \cdot e^{- 0.18578643} + 2 e^{4 (- 0.0464466)}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- 0.0464466) = 0.02866434 - 0.74314575 \cdot \frac{1}{e^{0.18578643}} + 2 e^{4 (- 0.0464466)}$

Combine $- 0.74314575$ and $\frac{1}{e^{0.18578643}}$ .

$f'' (- 0.0464466) = 0.02866434 + \frac{- 0.74314575}{e^{0.18578643}} + 2 e^{4 (- 0.0464466)}$

Move the negative in front of the fraction.

$f'' (- 0.0464466) = 0.02866434 - \frac{0.74314575}{e^{0.18578643}} + 2 e^{4 (- 0.0464466)}$

Replace $e$ with an approximation.

$f'' (- 0.0464466) = 0.02866434 - \frac{0.74314575}{{2.71828182}^{0.18578643}} + 2 e^{4 (- 0.0464466)}$

Raise $2.71828182$ to the power of $0.18578643$ .

$f'' (- 0.0464466) = 0.02866434 - \frac{0.74314575}{1.20416506} + 2 e^{4 (- 0.0464466)}$

Divide $0.74314575$ by $1.20416506$ .

$f'' (- 0.0464466) = 0.02866434 - 1 \cdot 0.61714607 + 2 e^{4 (- 0.0464466)}$

Multiply $- 1$ by $0.61714607$ .

$f'' (- 0.0464466) = 0.02866434 - 0.61714607 + 2 e^{4 (- 0.0464466)}$

Multiply $4$ by $- 0.0464466$ .

$f'' (- 0.0464466) = 0.02866434 - 0.61714607 + 2 e^{- 0.18578643}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (- 0.0464466) = 0.02866434 - 0.61714607 + 2 (\frac{1}{e^{0.18578643}})$

Combine $2$ and $\frac{1}{e^{0.18578643}}$ .

$f'' (- 0.0464466) = 0.02866434 - 0.61714607 + \frac{2}{e^{0.18578643}}$

Simplify by adding terms.

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Subtract $0.61714607$ from $0.02866434$ .

$f'' (- 0.0464466) = - 0.58848173 + \frac{2}{e^{0.18578643}}$

Add $- 0.58848173$ and $\frac{2}{e^{0.18578643}}$ .

$f'' (- 0.0464466) = 1.07242012$

The final answer is $1.07242012$ .

$1.07242012$

At $- 0.0464466$ , the second derivative is $1.07242012$ . Since this is positive, the second derivative is increasing on the interval $(- 0.1464466, \infty)$ .

Increasing on $(- 0.1464466, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- 0.85355339, 0.02397106), (- 0.1464466, 0.01193863)$ .

$(- 0.85355339, 0.02397106), (- 0.1464466, 0.01193863)$

Step 9