Calculus Examples

$x^{2} e^{9 x}$

Step 1

Write $x^{2} e^{9 x}$ as a function.

$f (x) = x^{2} e^{9 x}$

Step 2

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 $g (x) = e^{9 x}$ .

$f' (x) = x^{2} \frac{d}{d x} (e^{9 x}) + e^{9 x} \frac{d}{d x} (x^{2})$

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 $g (x) = 9 x$ .

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To apply the Chain Rule, set $u$ as $9 x$ .

$f' (x) = x^{2} (\frac{d}{d u} (e^{u}) \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x^{2})$

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

$f' (x) = x^{2} (e^{u} \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x^{2})$

Replace all occurrences of $u$ with $9 x$ .

$f' (x) = x^{2} (e^{9 x} \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x^{2})$

Differentiate.

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Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$f' (x) = x^{2} (e^{9 x} (9 \frac{d}{d x} (x))) + e^{9 x} \frac{d}{d x} (x^{2})$

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

$f' (x) = x^{2} (e^{9 x} (9 \cdot 1)) + e^{9 x} \frac{d}{d x} (x^{2})$

Simplify the expression.

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Multiply $9$ by $1$ .

$f' (x) = x^{2} (e^{9 x} \cdot 9) + e^{9 x} \frac{d}{d x} (x^{2})$

Move $9$ to the left of $e^{9 x}$ .

$f' (x) = x^{2} (9 \cdot e^{9 x}) + e^{9 x} \frac{d}{d x} (x^{2})$

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

$f' (x) = x^{2} (9 e^{9 x}) + e^{9 x} (2 x)$

Simplify.

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Reorder terms.

$f' (x) = 9 e^{9 x} x^{2} + 2 e^{9 x} x$

Reorder factors in $9 e^{9 x} x^{2} + 2 e^{9 x} x$ .

$f' (x) = 9 x^{2} e^{9 x} + 2 x e^{9 x}$

Find the second derivative.

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By the Sum Rule, the derivative of $9 x^{2} e^{9 x} + 2 x e^{9 x}$ with respect to $x$ is $\frac{d}{d x} [9 x^{2} e^{9 x}] + \frac{d}{d x} [2 x e^{9 x}]$ .

$f'' (x) = \frac{d}{d x} (9 x^{2} e^{9 x}) + \frac{d}{d x} (2 x e^{9 x})$

Evaluate $\frac{d}{d x} [9 x^{2} e^{9 x}]$ .

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Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2} e^{9 x}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2} e^{9 x}]$ .

$f'' (x) = 9 \frac{d}{d x} (x^{2} e^{9 x}) + \frac{d}{d x} (2 x e^{9 x})$

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 $g (x) = e^{9 x}$ .

$f'' (x) = 9 (x^{2} \frac{d}{d x} (e^{9 x}) + e^{9 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{9 x})$

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 $g (x) = 9 x$ .

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To apply the Chain Rule, set $u_{1}$ as $9 x$ .

$f'' (x) = 9 (x^{2} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{9 x})$

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

$f'' (x) = 9 (x^{2} (e^{u_{1}} \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{9 x})$

Replace all occurrences of $u_{1}$ with $9 x$ .

$f'' (x) = 9 (x^{2} (e^{9 x} \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{9 x})$

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$f'' (x) = 9 (x^{2} (e^{9 x} (9 \frac{d}{d x} (x))) + e^{9 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{9 x})$

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

$f'' (x) = 9 (x^{2} (e^{9 x} (9 \cdot 1)) + e^{9 x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{9 x})$

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

$f'' (x) = 9 (x^{2} (e^{9 x} (9 \cdot 1)) + e^{9 x} (2 x)) + \frac{d}{d x} (2 x e^{9 x})$

Multiply $9$ by $1$ .

$f'' (x) = 9 (x^{2} (e^{9 x} \cdot 9) + e^{9 x} (2 x)) + \frac{d}{d x} (2 x e^{9 x})$

Move $9$ to the left of $e^{9 x}$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + \frac{d}{d x} (2 x e^{9 x})$

Evaluate $\frac{d}{d x} [2 x e^{9 x}]$ .

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Since $2$ is constant with respect to $x$ , the derivative of $2 x e^{9 x}$ with respect to $x$ is $2 \frac{d}{d x} [x e^{9 x}]$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 \frac{d}{d x} (x e^{9 x})$

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 $g (x) = e^{9 x}$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x \frac{d}{d x} (e^{9 x}) + e^{9 x} \frac{d}{d x} (x))$

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 $g (x) = 9 x$ .

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To apply the Chain Rule, set $u_{2}$ as $9 x$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x))$

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

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (e^{u_{2}} \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x))$

Replace all occurrences of $u_{2}$ with $9 x$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (e^{9 x} \frac{d}{d x} (9 x)) + e^{9 x} \frac{d}{d x} (x))$

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (e^{9 x} (9 \frac{d}{d x} (x))) + e^{9 x} \frac{d}{d x} (x))$

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

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (e^{9 x} (9 \cdot 1)) + e^{9 x} \frac{d}{d x} (x))$

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

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (e^{9 x} (9 \cdot 1)) + e^{9 x} \cdot 1)$

Multiply $9$ by $1$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (e^{9 x} \cdot 9) + e^{9 x} \cdot 1)$

Move $9$ to the left of $e^{9 x}$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (9 \cdot e^{9 x}) + e^{9 x} \cdot 1)$

Multiply $e^{9 x}$ by $1$ .

$f'' (x) = 9 (x^{2} (9 e^{9 x}) + e^{9 x} (2 x)) + 2 (x (9 e^{9 x}) + e^{9 x})$

Simplify.

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Apply the distributive property.

$f'' (x) = 9 (x^{2} (9 e^{9 x})) + 9 (e^{9 x} (2 x)) + 2 (x (9 e^{9 x}) + e^{9 x})$

Apply the distributive property.

$f'' (x) = 9 (x^{2} (9 e^{9 x})) + 9 (e^{9 x} (2 x)) + 2 (x (9 e^{9 x})) + 2 e^{9 x}$

Combine terms.

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Multiply $9$ by $9$ .

$f'' (x) = 81 (x^{2} (e^{9 x})) + 9 (e^{9 x} (2 x)) + 2 (x (9 e^{9 x})) + 2 e^{9 x}$

Multiply $2$ by $9$ .

$f'' (x) = 81 x^{2} e^{9 x} + 18 (e^{9 x} (x)) + 2 (x (9 e^{9 x})) + 2 e^{9 x}$

Multiply $9$ by $2$ .

$f'' (x) = 81 x^{2} e^{9 x} + 18 e^{9 x} x + 18 (x (e^{9 x})) + 2 e^{9 x}$

Add $18 e^{9 x} x$ and $18 x e^{9 x}$ .

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Move $e^{9 x}$ .

$f'' (x) = 81 x^{2} e^{9 x} + 18 x e^{9 x} + 18 x e^{9 x} + 2 e^{9 x}$

Add $18 x e^{9 x}$ and $18 x e^{9 x}$ .

$f'' (x) = 81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x}$

Reorder terms.

$f'' (x) = 81 e^{9 x} x^{2} + 36 e^{9 x} x + 2 e^{9 x}$

Reorder factors in $81 e^{9 x} x^{2} + 36 e^{9 x} x + 2 e^{9 x}$ .

$f'' (x) = 81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x}$

The second derivative of $f (x)$ with respect to $x$ is $81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x}$ .

$81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x}$

Step 3

Set the second derivative equal to $0$ then solve the equation $81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x} = 0$ .

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Set the second derivative equal to $0$ .

$81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x} = 0$

Factor $e^{9 x}$ out of $81 x^{2} e^{9 x} + 36 x e^{9 x} + 2 e^{9 x}$ .

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Factor $e^{9 x}$ out of $81 x^{2} e^{9 x}$ .

$e^{9 x} (81 x^{2}) + 36 x e^{9 x} + 2 e^{9 x} = 0$

Factor $e^{9 x}$ out of $36 x e^{9 x}$ .

$e^{9 x} (81 x^{2}) + e^{9 x} (36 x) + 2 e^{9 x} = 0$

Factor $e^{9 x}$ out of $2 e^{9 x}$ .

$e^{9 x} (81 x^{2}) + e^{9 x} (36 x) + e^{9 x} \cdot 2 = 0$

Factor $e^{9 x}$ out of $e^{9 x} (81 x^{2}) + e^{9 x} (36 x)$ .

$e^{9 x} (81 x^{2} + 36 x) + e^{9 x} \cdot 2 = 0$

Factor $e^{9 x}$ out of $e^{9 x} (81 x^{2} + 36 x) + e^{9 x} \cdot 2$ .

$e^{9 x} (81 x^{2} + 36 x + 2) = 0$

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

$e^{9 x} = 0$

$81 x^{2} + 36 x + 2 = 0$

Set $e^{9 x}$ equal to $0$ and solve for $x$ .

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Set $e^{9 x}$ equal to $0$ .

$e^{9 x} = 0$

Solve $e^{9 x} = 0$ for $x$ .

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

$\ln (e^{9 x}) = \ln (0)$

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

Undefined

There is no solution for $e^{9 x} = 0$

No solution

Set $81 x^{2} + 36 x + 2$ equal to $0$ and solve for $x$ .

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Set $81 x^{2} + 36 x + 2$ equal to $0$ .

$81 x^{2} + 36 x + 2 = 0$

Solve $81 x^{2} + 36 x + 2 = 0$ 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 $a = 81$ , $b = 36$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{- 36 \pm \sqrt{36^{2} - 4 \cdot (81 \cdot 2)}}{2 \cdot 81}$

Simplify.

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Simplify the numerator.

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Raise $36$ to the power of $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 4 \cdot 81 \cdot 2}}{2 \cdot 81}$

Multiply $- 4 \cdot 81 \cdot 2$ .

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Multiply $- 4$ by $81$ .

$x = \frac{- 36 \pm \sqrt{1296 - 324 \cdot 2}}{2 \cdot 81}$

Multiply $- 324$ by $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 648}}{2 \cdot 81}$

Subtract $648$ from $1296$ .

$x = \frac{- 36 \pm \sqrt{648}}{2 \cdot 81}$

Rewrite $648$ as $18^{2} \cdot 2$ .

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Factor $324$ out of $648$ .

$x = \frac{- 36 \pm \sqrt{324 (2)}}{2 \cdot 81}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{- 36 \pm \sqrt{18^{2} \cdot 2}}{2 \cdot 81}$

Pull terms out from under the radical.

$x = \frac{- 36 \pm 18 \sqrt{2}}{2 \cdot 81}$

Multiply $2$ by $81$ .

$x = \frac{- 36 \pm 18 \sqrt{2}}{162}$

Simplify $\frac{- 36 \pm 18 \sqrt{2}}{162}$ .

$x = \frac{- 2 \pm \sqrt{2}}{9}$

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

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Simplify the numerator.

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Raise $36$ to the power of $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 4 \cdot 81 \cdot 2}}{2 \cdot 81}$

Multiply $- 4 \cdot 81 \cdot 2$ .

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Multiply $- 4$ by $81$ .

$x = \frac{- 36 \pm \sqrt{1296 - 324 \cdot 2}}{2 \cdot 81}$

Multiply $- 324$ by $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 648}}{2 \cdot 81}$

Subtract $648$ from $1296$ .

$x = \frac{- 36 \pm \sqrt{648}}{2 \cdot 81}$

Rewrite $648$ as $18^{2} \cdot 2$ .

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Factor $324$ out of $648$ .

$x = \frac{- 36 \pm \sqrt{324 (2)}}{2 \cdot 81}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{- 36 \pm \sqrt{18^{2} \cdot 2}}{2 \cdot 81}$

Pull terms out from under the radical.

$x = \frac{- 36 \pm 18 \sqrt{2}}{2 \cdot 81}$

Multiply $2$ by $81$ .

$x = \frac{- 36 \pm 18 \sqrt{2}}{162}$

Simplify $\frac{- 36 \pm 18 \sqrt{2}}{162}$ .

$x = \frac{- 2 \pm \sqrt{2}}{9}$

Change the $\pm$ to $+$ .

$x = \frac{- 2 + \sqrt{2}}{9}$

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

$x = \frac{- 1 \cdot 2 + \sqrt{2}}{9}$

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

$x = \frac{- 1 \cdot 2 - 1 (- \sqrt{2})}{9}$

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

$x = \frac{- 1 (2 - \sqrt{2})}{9}$

Move the negative in front of the fraction.

$x = - \frac{2 - \sqrt{2}}{9}$

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

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Simplify the numerator.

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Raise $36$ to the power of $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 4 \cdot 81 \cdot 2}}{2 \cdot 81}$

Multiply $- 4 \cdot 81 \cdot 2$ .

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Multiply $- 4$ by $81$ .

$x = \frac{- 36 \pm \sqrt{1296 - 324 \cdot 2}}{2 \cdot 81}$

Multiply $- 324$ by $2$ .

$x = \frac{- 36 \pm \sqrt{1296 - 648}}{2 \cdot 81}$

Subtract $648$ from $1296$ .

$x = \frac{- 36 \pm \sqrt{648}}{2 \cdot 81}$

Rewrite $648$ as $18^{2} \cdot 2$ .

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Factor $324$ out of $648$ .

$x = \frac{- 36 \pm \sqrt{324 (2)}}{2 \cdot 81}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{- 36 \pm \sqrt{18^{2} \cdot 2}}{2 \cdot 81}$

Pull terms out from under the radical.

$x = \frac{- 36 \pm 18 \sqrt{2}}{2 \cdot 81}$

Multiply $2$ by $81$ .

$x = \frac{- 36 \pm 18 \sqrt{2}}{162}$

Simplify $\frac{- 36 \pm 18 \sqrt{2}}{162}$ .

$x = \frac{- 2 \pm \sqrt{2}}{9}$

Change the $\pm$ to $-$ .

$x = \frac{- 2 - \sqrt{2}}{9}$

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

$x = \frac{- 1 \cdot 2 - \sqrt{2}}{9}$

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

$x = \frac{- 1 \cdot 2 - (\sqrt{2})}{9}$

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

$x = \frac{- 1 (2 + \sqrt{2})}{9}$

Move the negative in front of the fraction.

$x = - \frac{2 + \sqrt{2}}{9}$

The final answer is the combination of both solutions.

$x = - \frac{2 - \sqrt{2}}{9}, - \frac{2 + \sqrt{2}}{9}$

The final solution is all the values that make $e^{9 x} (81 x^{2} + 36 x + 2) = 0$ true.

$x = - \frac{2 - \sqrt{2}}{9}, - \frac{2 + \sqrt{2}}{9}$

Step 4

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

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Substitute $- 0.06508738$ in $f (x) = x^{2} e^{9 x}$ to find the value of $y$ .

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Replace the variable $x$ with $- 0.06508738$ in the expression.

$f (- 0.06508738) = {(- 0.06508738)}^{2} e^{9 (- 0.06508738)}$

Simplify the result.

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Raise $- 0.06508738$ to the power of $2$ .

$f (- 0.06508738) = 0.00423636 e^{9 (- 0.06508738)}$

Multiply $9$ by $- 0.06508738$ .

$f (- 0.06508738) = 0.00423636 e^{- 0.58578643}$

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

$f (- 0.06508738) = 0.00423636 (\frac{1}{e^{0.58578643}})$

Combine $0.00423636$ and $\frac{1}{e^{0.58578643}}$ .

$f (- 0.06508738) = \frac{0.00423636}{e^{0.58578643}}$

Replace $e$ with an approximation.

$f (- 0.06508738) = \frac{0.00423636}{{2.71828182}^{0.58578643}}$

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

$f (- 0.06508738) = \frac{0.00423636}{1.79640318}$

Divide $0.00423636$ by $1.79640318$ .

$f (- 0.06508738) = 0.00235824$

The final answer is $0.00235824$ .

$0.00235824$

The point found by substituting $- 0.06508738$ in $f (x) = x^{2} e^{9 x}$ is $(- 0.06508738, 0.00235824)$ . This point can be an inflection point.

$(- 0.06508738, 0.00235824)$

Substitute $- 0.37935706$ in $f (x) = x^{2} e^{9 x}$ to find the value of $y$ .

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Replace the variable $x$ with $- 0.37935706$ in the expression.

$f (- 0.37935706) = {(- 0.37935706)}^{2} e^{9 (- 0.37935706)}$

Simplify the result.

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Raise $- 0.37935706$ to the power of $2$ .

$f (- 0.37935706) = 0.14391178 e^{9 (- 0.37935706)}$

Multiply $9$ by $- 0.37935706$ .

$f (- 0.37935706) = 0.14391178 e^{- 3.41421356}$

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

$f (- 0.37935706) = 0.14391178 (\frac{1}{e^{3.41421356}})$

Combine $0.14391178$ and $\frac{1}{e^{3.41421356}}$ .

$f (- 0.37935706) = \frac{0.14391178}{e^{3.41421356}}$

Replace $e$ with an approximation.

$f (- 0.37935706) = \frac{0.14391178}{{2.71828182}^{3.41421356}}$

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

$f (- 0.37935706) = \frac{0.14391178}{30.39303779}$

Divide $0.14391178$ by $30.39303779$ .

$f (- 0.37935706) = 0.00473502$

The final answer is $0.00473502$ .

$0.00473502$

The point found by substituting $- 0.37935706$ in $f (x) = x^{2} e^{9 x}$ is $(- 0.37935706, 0.00473502)$ . This point can be an inflection point.

$(- 0.37935706, 0.00473502)$

Determine the points that could be inflection points.

$(- 0.06508738, 0.00235824), (- 0.37935706, 0.00473502)$

Step 5

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

$(- \infty, - 0.37935706) \cup (- 0.37935706, - 0.06508738) \cup (- 0.06508738, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 0.37935706)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.47935706$ in the expression.

$f'' (- 0.47935706) = 81 {(- 0.47935706)}^{2} e^{9 (- 0.47935706)} + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Simplify the result.

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Simplify each term.

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Raise $- 0.47935706$ to the power of $2$ .

$f'' (- 0.47935706) = 81 \cdot (0.22978319 e^{9 (- 0.47935706)}) + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Multiply $81$ by $0.22978319$ .

$f'' (- 0.47935706) = 18.61243866 e^{9 (- 0.47935706)} + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Multiply $9$ by $- 0.47935706$ .

$f'' (- 0.47935706) = 18.61243866 e^{- 4.31421356} + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

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

$f'' (- 0.47935706) = 18.61243866 (\frac{1}{e^{4.31421356}}) + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Combine $18.61243866$ and $\frac{1}{e^{4.31421356}}$ .

$f'' (- 0.47935706) = \frac{18.61243866}{e^{4.31421356}} + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Replace $e$ with an approximation.

$f'' (- 0.47935706) = \frac{18.61243866}{{2.71828182}^{4.31421356}} + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Raise $2.71828182$ to the power of $4.31421356$ .

$f'' (- 0.47935706) = \frac{18.61243866}{74.75481032} + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Divide $18.61243866$ by $74.75481032$ .

$f'' (- 0.47935706) = 0.24897981 + 36 (- 0.47935706) \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Multiply $36$ by $- 0.47935706$ .

$f'' (- 0.47935706) = 0.24897981 - 17.25685424 \cdot e^{9 (- 0.47935706)} + 2 e^{9 (- 0.47935706)}$

Multiply $9$ by $- 0.47935706$ .

$f'' (- 0.47935706) = 0.24897981 - 17.25685424 \cdot e^{- 4.31421356} + 2 e^{9 (- 0.47935706)}$

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

$f'' (- 0.47935706) = 0.24897981 - 17.25685424 \cdot \frac{1}{e^{4.31421356}} + 2 e^{9 (- 0.47935706)}$

Combine $- 17.25685424$ and $\frac{1}{e^{4.31421356}}$ .

$f'' (- 0.47935706) = 0.24897981 + \frac{- 17.25685424}{e^{4.31421356}} + 2 e^{9 (- 0.47935706)}$

Move the negative in front of the fraction.

$f'' (- 0.47935706) = 0.24897981 - \frac{17.25685424}{e^{4.31421356}} + 2 e^{9 (- 0.47935706)}$

Replace $e$ with an approximation.

$f'' (- 0.47935706) = 0.24897981 - \frac{17.25685424}{{2.71828182}^{4.31421356}} + 2 e^{9 (- 0.47935706)}$

Raise $2.71828182$ to the power of $4.31421356$ .

$f'' (- 0.47935706) = 0.24897981 - \frac{17.25685424}{74.75481032} + 2 e^{9 (- 0.47935706)}$

Divide $17.25685424$ by $74.75481032$ .

$f'' (- 0.47935706) = 0.24897981 - 1 \cdot 0.23084607 + 2 e^{9 (- 0.47935706)}$

Multiply $- 1$ by $0.23084607$ .

$f'' (- 0.47935706) = 0.24897981 - 0.23084607 + 2 e^{9 (- 0.47935706)}$

Multiply $9$ by $- 0.47935706$ .

$f'' (- 0.47935706) = 0.24897981 - 0.23084607 + 2 e^{- 4.31421356}$

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

$f'' (- 0.47935706) = 0.24897981 - 0.23084607 + 2 (\frac{1}{e^{4.31421356}})$

Combine $2$ and $\frac{1}{e^{4.31421356}}$ .

$f'' (- 0.47935706) = 0.24897981 - 0.23084607 + \frac{2}{e^{4.31421356}}$

Simplify by adding terms.

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Subtract $0.23084607$ from $0.24897981$ .

$f'' (- 0.47935706) = 0.01813374 + \frac{2}{e^{4.31421356}}$

Add $0.01813374$ and $\frac{2}{e^{4.31421356}}$ .

$f'' (- 0.47935706) = 0.04488787$

The final answer is $0.04488787$ .

$0.04488787$

At $- 0.47935706$ , the second derivative is $0.04488787$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 0.37935706)$ .

Increasing on $(- \infty, - 0.37935706)$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(- 0.37935706, - 0.06508738)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 0.\overline{2}$ in the expression.

$f'' (- 0.\overline{2}) = 81 {(- 0.\overline{2})}^{2} e^{9 (- 0.\overline{2})} + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Simplify the result.

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Simplify each term.

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Raise $- 0.\overline{2}$ to the power of $2$ .

$f'' (- 0.\overline{2}) = 81 \cdot (0.04938271 e^{9 (- 0.\overline{2})}) + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Multiply $81$ by $0.04938271$ .

$f'' (- 0.\overline{2}) = 4 e^{9 (- 0.\overline{2})} + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Multiply $9$ by $- 0.\overline{2}$ .

$f'' (- 0.\overline{2}) = 4 e^{- 2} + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

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

$f'' (- 0.\overline{2}) = 4 (\frac{1}{e^{2}}) + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

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

$f'' (- 0.\overline{2}) = \frac{4}{e^{2}} + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Replace $e$ with an approximation.

$f'' (- 0.\overline{2}) = \frac{4}{{2.71828182}^{2}} + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Raise $2.71828182$ to the power of $2$ .

$f'' (- 0.\overline{2}) = \frac{4}{7.38905609} + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Divide $4$ by $7.38905609$ .

$f'' (- 0.\overline{2}) = 0.54134113 + 36 (- 0.\overline{2}) \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Multiply $36$ by $- 0.\overline{2}$ .

$f'' (- 0.\overline{2}) = 0.54134113 - 8 \cdot e^{9 (- 0.\overline{2})} + 2 e^{9 (- 0.\overline{2})}$

Multiply $9$ by $- 0.\overline{2}$ .

$f'' (- 0.\overline{2}) = 0.54134113 - 8 \cdot e^{- 2} + 2 e^{9 (- 0.\overline{2})}$

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

$f'' (- 0.\overline{2}) = 0.54134113 - 8 \cdot \frac{1}{e^{2}} + 2 e^{9 (- 0.\overline{2})}$

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

$f'' (- 0.\overline{2}) = 0.54134113 + \frac{- 8}{e^{2}} + 2 e^{9 (- 0.\overline{2})}$

Move the negative in front of the fraction.

$f'' (- 0.\overline{2}) = 0.54134113 - \frac{8}{e^{2}} + 2 e^{9 (- 0.\overline{2})}$

Replace $e$ with an approximation.

$f'' (- 0.\overline{2}) = 0.54134113 - \frac{8}{{2.71828182}^{2}} + 2 e^{9 (- 0.\overline{2})}$

Raise $2.71828182$ to the power of $2$ .

$f'' (- 0.\overline{2}) = 0.54134113 - \frac{8}{7.38905609} + 2 e^{9 (- 0.\overline{2})}$

Divide $8$ by $7.38905609$ .

$f'' (- 0.\overline{2}) = 0.54134113 - 1 \cdot 1.08268226 + 2 e^{9 (- 0.\overline{2})}$

Multiply $- 1$ by $1.08268226$ .

$f'' (- 0.\overline{2}) = 0.54134113 - 1.08268226 + 2 e^{9 (- 0.\overline{2})}$

Multiply $9$ by $- 0.\overline{2}$ .

$f'' (- 0.\overline{2}) = 0.54134113 - 1.08268226 + 2 e^{- 2}$

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

$f'' (- 0.\overline{2}) = 0.54134113 - 1.08268226 + 2 (\frac{1}{e^{2}})$

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

$f'' (- 0.\overline{2}) = 0.54134113 - 1.08268226 + \frac{2}{e^{2}}$

Simplify by adding terms.

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Subtract $1.08268226$ from $0.54134113$ .

$f'' (- 0.\overline{2}) = - 0.54134113 + \frac{2}{e^{2}}$

Add $- 0.54134113$ and $\frac{2}{e^{2}}$ .

$f'' (- 0.\overline{2}) = - 0.27067056$

The final answer is $- 0.27067056$ .

$- 0.27067056$

At $- 0.\overline{2}$ , the second derivative is $- 0.27067056$ . Since this is negative, the second derivative is decreasing on the interval $(- 0.37935706, - 0.06508738)$

Decreasing on $(- 0.37935706, - 0.06508738)$ since $f'' (x) < 0$

Step 8

Substitute a value from the interval $(- 0.06508738, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $0.03491261$ in the expression.

$f'' (0.03491261) = 81 {(0.03491261)}^{2} e^{9 (0.03491261)} + 36 (0.03491261) \cdot e^{9 (0.03491261)} + 2 e^{9 (0.03491261)}$

Simplify the result.

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Simplify each term.

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Raise $0.03491261$ to the power of $2$ .

$f'' (0.03491261) = 81 \cdot (0.00121889 e^{9 (0.03491261)}) + 36 (0.03491261) \cdot e^{9 (0.03491261)} + 2 e^{9 (0.03491261)}$

Multiply $81$ by $0.00121889$ .

$f'' (0.03491261) = 0.09873016 e^{9 (0.03491261)} + 36 (0.03491261) \cdot e^{9 (0.03491261)} + 2 e^{9 (0.03491261)}$

Multiply $9$ by $0.03491261$ .

$f'' (0.03491261) = 0.09873016 e^{0.31421356} + 36 (0.03491261) \cdot e^{9 (0.03491261)} + 2 e^{9 (0.03491261)}$

Multiply $0.09873016$ by $e^{0.31421356}$ .

$f'' (0.03491261) = 0.13517957 + 36 (0.03491261) \cdot e^{9 (0.03491261)} + 2 e^{9 (0.03491261)}$

Multiply $36$ by $0.03491261$ .

$f'' (0.03491261) = 0.13517957 + 1.25685424 \cdot e^{9 (0.03491261)} + 2 e^{9 (0.03491261)}$

Multiply $9$ by $0.03491261$ .

$f'' (0.03491261) = 0.13517957 + 1.25685424 \cdot e^{0.31421356} + 2 e^{9 (0.03491261)}$

Multiply $1.25685424$ by $e^{0.31421356}$ .

$f'' (0.03491261) = 0.13517957 + 1.72086235 + 2 e^{9 (0.03491261)}$

Multiply $9$ by $0.03491261$ .

$f'' (0.03491261) = 0.13517957 + 1.72086235 + 2 e^{0.31421356}$

Simplify by adding terms.

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Add $0.13517957$ and $1.72086235$ .

$f'' (0.03491261) = 1.85604192 + 2 e^{0.31421356}$

Add $1.85604192$ and $2 e^{0.31421356}$ .

$f'' (0.03491261) = 4.59440614$

The final answer is $4.59440614$ .

$4.59440614$

At $0.03491261$ , the second derivative is $4.59440614$ . Since this is positive, the second derivative is increasing on the interval $(- 0.06508738, \infty)$ .

Increasing on $(- 0.06508738, \infty)$ since $f'' (x) > 0$

Step 9

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 $(- 0.37935706, 0.00473502), (- 0.06508738, 0.00235824)$ .

$(- 0.37935706, 0.00473502), (- 0.06508738, 0.00235824)$

Step 10