Calculus Examples

$f (t) = e^{- 2 t^{2}}$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 2 t^{2}$ .

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Step 1.1.1.1

To apply the Chain Rule, set $u$ as $- 2 t^{2}$ .

$f' (t) = \frac{d}{d u} (e^{u}) \frac{d}{d t} (- 2 t^{2})$

Step 1.1.1.2

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

$f' (t) = e^{u} \frac{d}{d t} (- 2 t^{2})$

Step 1.1.1.3

Replace all occurrences of $u$ with $- 2 t^{2}$ .

$f' (t) = e^{- 2 t^{2}} \frac{d}{d t} (- 2 t^{2})$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

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

$f' (t) = e^{- 2 t^{2}} (- 2 \frac{d}{d t} t^{2})$

Step 1.1.2.2

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

$f' (t) = e^{- 2 t^{2}} (- 2 (2 t))$

Step 1.1.2.3

Multiply $2$ by $- 2$ .

$f' (t) = e^{- 2 t^{2}} (- 4 t)$

Step 1.1.3

Simplify.

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Step 1.1.3.1

Reorder the factors of $e^{- 2 t^{2}} (- 4 t)$ .

$f' (t) = - 4 e^{- 2 t^{2}} t$

Step 1.1.3.2

Reorder factors in $- 4 e^{- 2 t^{2}} t$ .

$f' (t) = - 4 t e^{- 2 t^{2}}$

Step 1.2

Find the second derivative.

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Step 1.2.1

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

$f'' (t) = - 4 \frac{d}{d t} (t e^{- 2 t^{2}})$

Step 1.2.2

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

$f'' (t) = - 4 (t \frac{d}{d t} (e^{- 2 t^{2}}) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 2 t^{2}$ .

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Step 1.2.3.1

To apply the Chain Rule, set $u$ as $- 2 t^{2}$ .

$f'' (t) = - 4 (t (\frac{d}{d u} (e^{u}) \frac{d}{d t} (- 2 t^{2})) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.3.2

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

$f'' (t) = - 4 (t (e^{u} \frac{d}{d t} (- 2 t^{2})) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.3.3

Replace all occurrences of $u$ with $- 2 t^{2}$ .

$f'' (t) = - 4 (t (e^{- 2 t^{2}} \frac{d}{d t} (- 2 t^{2})) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.4

Differentiate.

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Step 1.2.4.1

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

$f'' (t) = - 4 (t (e^{- 2 t^{2}} (- 2 \frac{d}{d t} t^{2})) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.4.2

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

$f'' (t) = - 4 (t (e^{- 2 t^{2}} (- 2 (2 t))) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.4.3

Multiply $2$ by $- 2$ .

$f'' (t) = - 4 (t (e^{- 2 t^{2}} (- 4 t)) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.5

Raise $t$ to the power of $1$ .

$f'' (t) = - 4 (t \cdot t (e^{- 2 t^{2}} \cdot (- 4)) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.6

Raise $t$ to the power of $1$ .

$f'' (t) = - 4 (t \cdot t (e^{- 2 t^{2}} \cdot (- 4)) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (t) = - 4 (t^{1 + 1} (e^{- 2 t^{2}} \cdot (- 4)) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.8

Simplify the expression.

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Step 1.2.8.1

Add $1$ and $1$ .

$f'' (t) = - 4 (t^{2} (e^{- 2 t^{2}} \cdot (- 4)) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.8.2

Move $- 4$ to the left of $e^{- 2 t^{2}}$ .

$f'' (t) = - 4 (t^{2} (- 4 \cdot e^{- 2 t^{2}}) + e^{- 2 t^{2}} \frac{d}{d t} (t))$

Step 1.2.9

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

$f'' (t) = - 4 (t^{2} (- 4 e^{- 2 t^{2}}) + e^{- 2 t^{2}} \cdot 1)$

Step 1.2.10

Multiply $e^{- 2 t^{2}}$ by $1$ .

$f'' (t) = - 4 (t^{2} (- 4 e^{- 2 t^{2}}) + e^{- 2 t^{2}})$

Step 1.2.11

Simplify.

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Step 1.2.11.1

Apply the distributive property.

$f'' (t) = - 4 (t^{2} (- 4 e^{- 2 t^{2}})) - 4 e^{- 2 t^{2}}$

Step 1.2.11.2

Multiply $- 4$ by $- 4$ .

$f'' (t) = 16 t^{2} e^{- 2 t^{2}} - 4 e^{- 2 t^{2}}$

Step 1.2.11.3

Reorder terms.

$f'' (t) = 16 e^{- 2 t^{2}} t^{2} - 4 e^{- 2 t^{2}}$

Step 1.2.11.4

Reorder factors in $16 e^{- 2 t^{2}} t^{2} - 4 e^{- 2 t^{2}}$ .

$f'' (t) = 16 t^{2} e^{- 2 t^{2}} - 4 e^{- 2 t^{2}}$

Step 1.3

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

$16 t^{2} e^{- 2 t^{2}} - 4 e^{- 2 t^{2}}$

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

$16 t^{2} e^{- 2 t^{2}} - 4 e^{- 2 t^{2}} = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

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

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Step 2.2.1.1

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

$4 e^{- 2 t^{2}} (4 t^{2}) - 4 e^{- 2 t^{2}} = 0$

Step 2.2.1.2

Factor $4 e^{- 2 t^{2}}$ out of $- 4 e^{- 2 t^{2}}$ .

$4 e^{- 2 t^{2}} (4 t^{2}) + 4 e^{- 2 t^{2}} (- 1) = 0$

Step 2.2.1.3

Factor $4 e^{- 2 t^{2}}$ out of $4 e^{- 2 t^{2}} (4 t^{2}) + 4 e^{- 2 t^{2}} (- 1)$ .

$4 e^{- 2 t^{2}} (4 t^{2} - 1) = 0$

Step 2.2.2

Rewrite $4 t^{2}$ as ${(2 t)}^{2}$ .

$4 e^{- 2 t^{2}} ({(2 t)}^{2} - 1) = 0$

Step 2.2.3

Rewrite $1$ as $1^{2}$ .

$4 e^{- 2 t^{2}} ({(2 t)}^{2} - 1^{2}) = 0$

Step 2.2.4

Factor.

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Step 2.2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 t$ and $b = 1$ .

$4 e^{- 2 t^{2}} ((2 t + 1) (2 t - 1)) = 0$

Step 2.2.4.2

Remove unnecessary parentheses.

$4 e^{- 2 t^{2}} (2 t + 1) (2 t - 1) = 0$

Step 2.3

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

$e^{- 2 t^{2}} = 0$

$2 t + 1 = 0$

$2 t - 1 = 0$

Step 2.4

Set $e^{- 2 t^{2}}$ equal to $0$ and solve for $t$ .

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Step 2.4.1

Set $e^{- 2 t^{2}}$ equal to $0$ .

$e^{- 2 t^{2}} = 0$

Step 2.4.2

Solve $e^{- 2 t^{2}} = 0$ for $t$ .

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Step 2.4.2.1

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

$\ln (e^{- 2 t^{2}}) = \ln (0)$

Step 2.4.2.2

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

Undefined

Step 2.4.2.3

There is no solution for $e^{- 2 t^{2}} = 0$

No solution

Step 2.5

Set $2 t + 1$ equal to $0$ and solve for $t$ .

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Step 2.5.1

Set $2 t + 1$ equal to $0$ .

$2 t + 1 = 0$

Step 2.5.2

Solve $2 t + 1 = 0$ for $t$ .

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Step 2.5.2.1

Subtract $1$ from both sides of the equation.

$2 t = - 1$

Step 2.5.2.2

Divide each term in $2 t = - 1$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 t = - 1$ by $2$ .

$\frac{2 t}{2} = \frac{- 1}{2}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 t}{2} = \frac{- 1}{2}$

Step 2.5.2.2.2.1.2

Divide $t$ by $1$ .

$t = \frac{- 1}{2}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$t = - \frac{1}{2}$

Step 2.6

Set $2 t - 1$ equal to $0$ and solve for $t$ .

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Step 2.6.1

Set $2 t - 1$ equal to $0$ .

$2 t - 1 = 0$

Step 2.6.2

Solve $2 t - 1 = 0$ for $t$ .

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Step 2.6.2.1

Add $1$ to both sides of the equation.

$2 t = 1$

Step 2.6.2.2

Divide each term in $2 t = 1$ by $2$ and simplify.

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Step 2.6.2.2.1

Divide each term in $2 t = 1$ by $2$ .

$\frac{2 t}{2} = \frac{1}{2}$

Step 2.6.2.2.2

Simplify the left side.

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Step 2.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.6.2.2.2.1.1

Cancel the common factor.

$\frac{2 t}{2} = \frac{1}{2}$

Step 2.6.2.2.2.1.2

Divide $t$ by $1$ .

$t = \frac{1}{2}$

Step 2.7

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

$t = - \frac{1}{2}, \frac{1}{2}$

Step 3

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

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Step 3.1

Substitute $- \frac{1}{2}$ in $f (t) = e^{- 2 t^{2}}$ to find the value of $y$ .

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Step 3.1.1

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

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

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

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

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Step 3.1.2.1.1

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

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

Step 3.1.2.1.2

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

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

Step 3.1.2.2

Simplify the expression.

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Step 3.1.2.2.1

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

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

Step 3.1.2.2.2

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

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

Step 3.1.2.2.3

One to any power is one.

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

Step 3.1.2.2.4

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

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

Step 3.1.2.3

Cancel the common factor of $2$ .

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Step 3.1.2.3.1

Factor $2$ out of $- 2$ .

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

Step 3.1.2.3.2

Factor $2$ out of $4$ .

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

Step 3.1.2.3.3

Cancel the common factor.

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

Step 3.1.2.3.4

Rewrite the expression.

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

Step 3.1.2.4

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

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

Step 3.1.2.5

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

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

Step 3.1.2.6

The final answer is $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{1}{e^{\frac{1}{2}}}$

Step 3.2

The point found by substituting $- \frac{1}{2}$ in $f (t) = e^{- 2 t^{2}}$ is $(- \frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$ . This point can be an inflection point.

$(- \frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$

Step 3.3

Substitute $\frac{1}{2}$ in $f (t) = e^{- 2 t^{2}}$ to find the value of $y$ .

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Step 3.3.1

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

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

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify the expression.

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Step 3.3.2.1.1

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

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

Step 3.3.2.1.2

One to any power is one.

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

Step 3.3.2.1.3

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

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

Step 3.3.2.2

Cancel the common factor of $2$ .

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Step 3.3.2.2.1

Factor $2$ out of $- 2$ .

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

Step 3.3.2.2.2

Factor $2$ out of $4$ .

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

Step 3.3.2.2.3

Cancel the common factor.

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

Step 3.3.2.2.4

Rewrite the expression.

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

Step 3.3.2.3

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

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

Step 3.3.2.4

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

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

Step 3.3.2.5

The final answer is $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{1}{e^{\frac{1}{2}}}$

Step 3.4

The point found by substituting $\frac{1}{2}$ in $f (t) = e^{- 2 t^{2}}$ is $(\frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$ . This point can be an inflection point.

$(\frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$

Step 3.5

Determine the points that could be inflection points.

$(- \frac{1}{2}, \frac{1}{e^{\frac{1}{2}}}), (\frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$

Step 4

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

$(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $t$ with $- 0.6$ in the expression.

$f'' (- 0.6) = 16 {(- 0.6)}^{2} e^{- 2 {(- 0.6)}^{2}} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Raise $- 0.6$ to the power of $2$ .

$f'' (- 0.6) = 16 \cdot (0.36 e^{- 2 {(- 0.6)}^{2}}) - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.2

Multiply $16$ by $0.36$ .

$f'' (- 0.6) = 5.76 e^{- 2 {(- 0.6)}^{2}} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.3

Raise $- 0.6$ to the power of $2$ .

$f'' (- 0.6) = 5.76 e^{- 2 \cdot 0.36} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.4

Multiply $- 2$ by $0.36$ .

$f'' (- 0.6) = 5.76 e^{- 0.72} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.5

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

$f'' (- 0.6) = 5.76 (\frac{1}{e^{0.72}}) - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.6

Combine $5.76$ and $\frac{1}{e^{0.72}}$ .

$f'' (- 0.6) = \frac{5.76}{e^{0.72}} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.7

Replace $e$ with an approximation.

$f'' (- 0.6) = \frac{5.76}{{2.71828182}^{0.72}} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.8

Raise $2.71828182$ to the power of $0.72$ .

$f'' (- 0.6) = \frac{5.76}{2.05443321} - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.9

Divide $5.76$ by $2.05443321$ .

$f'' (- 0.6) = 2.80369299 - 4 e^{- 2 {(- 0.6)}^{2}}$

Step 5.2.1.10

Raise $- 0.6$ to the power of $2$ .

$f'' (- 0.6) = 2.80369299 - 4 e^{- 2 \cdot 0.36}$

Step 5.2.1.11

Multiply $- 2$ by $0.36$ .

$f'' (- 0.6) = 2.80369299 - 4 e^{- 0.72}$

Step 5.2.1.12

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

$f'' (- 0.6) = 2.80369299 - 4 \frac{1}{e^{0.72}}$

Step 5.2.1.13

Combine $- 4$ and $\frac{1}{e^{0.72}}$ .

$f'' (- 0.6) = 2.80369299 + \frac{- 4}{e^{0.72}}$

Step 5.2.1.14

Move the negative in front of the fraction.

$f'' (- 0.6) = 2.80369299 - \frac{4}{e^{0.72}}$

Step 5.2.2

Subtract $\frac{4}{e^{0.72}}$ from $2.80369299$ .

$f'' (- 0.6) = 0.85668397$

Step 5.2.3

The final answer is $0.85668397$ .

$0.85668397$

Step 5.3

At $- 0.6$ , the second derivative is $0.85668397$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - \frac{1}{2})$ .

Increasing on $(- \infty, - \frac{1}{2})$ since $f'' (t) > 0$

Step 6

Substitute a value from the interval $(- \frac{1}{2}, \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $t$ with $0$ in the expression.

$f'' (0) = 16 {(0)}^{2} e^{- 2 {(0)}^{2}} - 4 e^{- 2 {(0)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Raising $0$ to any positive power yields $0$ .

$f'' (0) = 16 \cdot (0 e^{- 2 {(0)}^{2}}) - 4 e^{- 2 {(0)}^{2}}$

Step 6.2.1.2

Multiply $16$ by $0$ .

$f'' (0) = 0 e^{- 2 {(0)}^{2}} - 4 e^{- 2 {(0)}^{2}}$

Step 6.2.1.3

Raising $0$ to any positive power yields $0$ .

$f'' (0) = 0 e^{- 2 \cdot 0} - 4 e^{- 2 {(0)}^{2}}$

Step 6.2.1.4

Multiply $- 2$ by $0$ .

$f'' (0) = 0 e^{0} - 4 e^{- 2 {(0)}^{2}}$

Step 6.2.1.5

Anything raised to $0$ is $1$ .

$f'' (0) = 0 \cdot 1 - 4 e^{- 2 {(0)}^{2}}$

Step 6.2.1.6

Multiply $0$ by $1$ .

$f'' (0) = 0 - 4 e^{- 2 {(0)}^{2}}$

Step 6.2.1.7

Raising $0$ to any positive power yields $0$ .

$f'' (0) = 0 - 4 e^{- 2 \cdot 0}$

Step 6.2.1.8

Multiply $- 2$ by $0$ .

$f'' (0) = 0 - 4 e^{0}$

Step 6.2.1.9

Anything raised to $0$ is $1$ .

$f'' (0) = 0 - 4 \cdot 1$

Step 6.2.1.10

Multiply $- 4$ by $1$ .

$f'' (0) = 0 - 4$

Step 6.2.2

Subtract $4$ from $0$ .

$f'' (0) = - 4$

Step 6.2.3

The final answer is $- 4$ .

$- 4$

Step 6.3

At $0$ , the second derivative is $- 4$ . Since this is negative, the second derivative is decreasing on the interval $(- \frac{1}{2}, \frac{1}{2})$

Decreasing on $(- \frac{1}{2}, \frac{1}{2})$ since $f'' (t) < 0$

Step 7

Substitute a value from the interval $(\frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $t$ with $0.6$ in the expression.

$f'' (0.6) = 16 {(0.6)}^{2} e^{- 2 {(0.6)}^{2}} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $0.6$ to the power of $2$ .

$f'' (0.6) = 16 \cdot (0.36 e^{- 2 {(0.6)}^{2}}) - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.2

Multiply $16$ by $0.36$ .

$f'' (0.6) = 5.76 e^{- 2 {(0.6)}^{2}} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.3

Raise $0.6$ to the power of $2$ .

$f'' (0.6) = 5.76 e^{- 2 \cdot 0.36} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.4

Multiply $- 2$ by $0.36$ .

$f'' (0.6) = 5.76 e^{- 0.72} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.5

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

$f'' (0.6) = 5.76 (\frac{1}{e^{0.72}}) - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.6

Combine $5.76$ and $\frac{1}{e^{0.72}}$ .

$f'' (0.6) = \frac{5.76}{e^{0.72}} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.7

Replace $e$ with an approximation.

$f'' (0.6) = \frac{5.76}{{2.71828182}^{0.72}} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.8

Raise $2.71828182$ to the power of $0.72$ .

$f'' (0.6) = \frac{5.76}{2.05443321} - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.9

Divide $5.76$ by $2.05443321$ .

$f'' (0.6) = 2.80369299 - 4 e^{- 2 {(0.6)}^{2}}$

Step 7.2.1.10

Raise $0.6$ to the power of $2$ .

$f'' (0.6) = 2.80369299 - 4 e^{- 2 \cdot 0.36}$

Step 7.2.1.11

Multiply $- 2$ by $0.36$ .

$f'' (0.6) = 2.80369299 - 4 e^{- 0.72}$

Step 7.2.1.12

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

$f'' (0.6) = 2.80369299 - 4 \frac{1}{e^{0.72}}$

Step 7.2.1.13

Combine $- 4$ and $\frac{1}{e^{0.72}}$ .

$f'' (0.6) = 2.80369299 + \frac{- 4}{e^{0.72}}$

Step 7.2.1.14

Move the negative in front of the fraction.

$f'' (0.6) = 2.80369299 - \frac{4}{e^{0.72}}$

Step 7.2.2

Subtract $\frac{4}{e^{0.72}}$ from $2.80369299$ .

$f'' (0.6) = 0.85668397$

Step 7.2.3

The final answer is $0.85668397$ .

$0.85668397$

Step 7.3

At $0.6$ , the second derivative is $0.85668397$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{2}, \infty)$ .

Increasing on $(\frac{1}{2}, \infty)$ since $f'' (t) > 0$

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 $(- \frac{1}{2}, \frac{1}{e^{\frac{1}{2}}}), (\frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$ .

$(- \frac{1}{2}, \frac{1}{e^{\frac{1}{2}}}), (\frac{1}{2}, \frac{1}{e^{\frac{1}{2}}})$

Step 9