Calculus Examples

$f (x) = e^{- 2.5 x^{2}}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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 $g (x) = - 2.5 x^{2}$ .

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Step 1.1.1.1.1

To apply the Chain Rule, set $u$ as $- 2.5 x^{2}$ .

$f' (x) = \frac{d}{d u} (e^{u}) \frac{d}{d x} (- 2.5 x^{2})$

Step 1.1.1.1.2

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

$f' (x) = e^{u} \frac{d}{d x} (- 2.5 x^{2})$

Step 1.1.1.1.3

Replace all occurrences of $u$ with $- 2.5 x^{2}$ .

$f' (x) = e^{- 2.5 x^{2}} \frac{d}{d x} (- 2.5 x^{2})$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

Since $- 2.5$ is constant with respect to $x$ , the derivative of $- 2.5 x^{2}$ with respect to $x$ is $- 2.5 \frac{d}{d x} [x^{2}]$ .

$f' (x) = e^{- 2.5 x^{2}} (- 2.5 \frac{d}{d x} x^{2})$

Step 1.1.1.2.2

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

$f' (x) = e^{- 2.5 x^{2}} (- 2.5 (2 x))$

Step 1.1.1.2.3

Multiply $2$ by $- 2.5$ .

$f' (x) = e^{- 2.5 x^{2}} (- 5 x)$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Reorder the factors of $e^{- 2.5 x^{2}} (- 5 x)$ .

$f' (x) = - 5 e^{- 2.5 x^{2}} x$

Step 1.1.1.3.2

Reorder factors in $- 5 e^{- 2.5 x^{2}} x$ .

$f' (x) = - 5 x e^{- 2.5 x^{2}}$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x e^{- 2.5 x^{2}}$ with respect to $x$ is $- 5 \frac{d}{d x} [x e^{- 2.5 x^{2}}]$ .

$f'' (x) = - 5 \frac{d}{d x} (x e^{- 2.5 x^{2}})$

Step 1.1.2.2

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 $g (x) = e^{- 2.5 x^{2}}$ .

$f'' (x) = - 5 (x \frac{d}{d x} (e^{- 2.5 x^{2}}) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.3

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 $g (x) = - 2.5 x^{2}$ .

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Step 1.1.2.3.1

To apply the Chain Rule, set $u$ as $- 2.5 x^{2}$ .

$f'' (x) = - 5 (x (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- 2.5 x^{2})) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.3.2

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

$f'' (x) = - 5 (x (e^{u} \frac{d}{d x} (- 2.5 x^{2})) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.3.3

Replace all occurrences of $u$ with $- 2.5 x^{2}$ .

$f'' (x) = - 5 (x (e^{- 2.5 x^{2}} \frac{d}{d x} (- 2.5 x^{2})) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.4

Differentiate.

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Step 1.1.2.4.1

Since $- 2.5$ is constant with respect to $x$ , the derivative of $- 2.5 x^{2}$ with respect to $x$ is $- 2.5 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 5 (x (e^{- 2.5 x^{2}} (- 2.5 \frac{d}{d x} x^{2})) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.4.2

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

$f'' (x) = - 5 (x (e^{- 2.5 x^{2}} (- 2.5 (2 x))) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.4.3

Multiply $2$ by $- 2.5$ .

$f'' (x) = - 5 (x (e^{- 2.5 x^{2}} (- 5 x)) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.5

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

$f'' (x) = - 5 (x \cdot x (e^{- 2.5 x^{2}} \cdot (- 5)) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.6

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

$f'' (x) = - 5 (x \cdot x (e^{- 2.5 x^{2}} \cdot (- 5)) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.7

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

$f'' (x) = - 5 (x^{1 + 1} (e^{- 2.5 x^{2}} \cdot (- 5)) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.8

Simplify the expression.

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Step 1.1.2.8.1

Add $1$ and $1$ .

$f'' (x) = - 5 (x^{2} (e^{- 2.5 x^{2}} \cdot (- 5)) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.8.2

Move $- 5$ to the left of $e^{- 2.5 x^{2}}$ .

$f'' (x) = - 5 (x^{2} (- 5 \cdot e^{- 2.5 x^{2}}) + e^{- 2.5 x^{2}} \frac{d}{d x} (x))$

Step 1.1.2.9

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

$f'' (x) = - 5 (x^{2} (- 5 e^{- 2.5 x^{2}}) + e^{- 2.5 x^{2}} \cdot 1)$

Step 1.1.2.10

Multiply $e^{- 2.5 x^{2}}$ by $1$ .

$f'' (x) = - 5 (x^{2} (- 5 e^{- 2.5 x^{2}}) + e^{- 2.5 x^{2}})$

Step 1.1.2.11

Simplify.

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Step 1.1.2.11.1

Apply the distributive property.

$f'' (x) = - 5 (x^{2} (- 5 e^{- 2.5 x^{2}})) - 5 e^{- 2.5 x^{2}}$

Step 1.1.2.11.2

Multiply $- 5$ by $- 5$ .

$f'' (x) = 25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}}$

Step 1.1.2.11.3

Reorder terms.

$f'' (x) = 25 e^{- 2.5 x^{2}} x^{2} - 5 e^{- 2.5 x^{2}}$

Step 1.1.2.11.4

Reorder factors in $25 e^{- 2.5 x^{2}} x^{2} - 5 e^{- 2.5 x^{2}}$ .

$f'' (x) = 25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}}$ .

$25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}} = 0$

Step 1.2.2

Factor the left side of the equation.

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Step 1.2.2.1

Factor $5 e^{- 2.5 x^{2}}$ out of $25 x^{2} e^{- 2.5 x^{2}} - 5 e^{- 2.5 x^{2}}$ .

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Step 1.2.2.1.1

Factor $5 e^{- 2.5 x^{2}}$ out of $25 x^{2} e^{- 2.5 x^{2}}$ .

$5 e^{- 2.5 x^{2}} (5 x^{2}) - 5 e^{- 2.5 x^{2}} = 0$

Step 1.2.2.1.2

Factor $5 e^{- 2.5 x^{2}}$ out of $- 5 e^{- 2.5 x^{2}}$ .

$5 e^{- 2.5 x^{2}} (5 x^{2}) + 5 e^{- 2.5 x^{2}} (- 1) = 0$

Step 1.2.2.1.3

Factor $5 e^{- 2.5 x^{2}}$ out of $5 e^{- 2.5 x^{2}} (5 x^{2}) + 5 e^{- 2.5 x^{2}} (- 1)$ .

$5 e^{- 2.5 x^{2}} (5 x^{2} - 1) = 0$

Step 1.2.2.2

Rewrite $5 x^{2}$ as ${(2.23606797 x)}^{2}$ .

$5 e^{- 2.5 x^{2}} ({(2.23606797 x)}^{2} - 1) = 0$

Step 1.2.2.3

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

$5 e^{- 2.5 x^{2}} ({(2.23606797 x)}^{2} - 1^{2}) = 0$

Step 1.2.2.4

Factor.

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Step 1.2.2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2.23606797 x$ and $b = 1$ .

$5 e^{- 2.5 x^{2}} ((2.23606797 x + 1) (2.23606797 x - 1)) = 0$

Step 1.2.2.4.2

Remove unnecessary parentheses.

$5 e^{- 2.5 x^{2}} (2.23606797 x + 1) (2.23606797 x - 1) = 0$

Step 1.2.3

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

$e^{- 2.5 x^{2}} = 0$

$2.23606797 x + 1 = 0$

$2.23606797 x - 1 = 0$

Step 1.2.4

Set $e^{- 2.5 x^{2}}$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $e^{- 2.5 x^{2}}$ equal to $0$ .

$e^{- 2.5 x^{2}} = 0$

Step 1.2.4.2

Solve $e^{- 2.5 x^{2}} = 0$ for $x$ .

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Step 1.2.4.2.1

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

$\ln (e^{- 2.5 x^{2}}) = \ln (0)$

Step 1.2.4.2.2

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

Undefined

Step 1.2.4.2.3

There is no solution for $e^{- 2.5 x^{2}} = 0$

No solution

Step 1.2.5

Set $2.23606797 x + 1$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $2.23606797 x + 1$ equal to $0$ .

$2.23606797 x + 1 = 0$

Step 1.2.5.2

Solve $2.23606797 x + 1 = 0$ for $x$ .

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Step 1.2.5.2.1

Subtract $1$ from both sides of the equation.

$2.23606797 x = - 1$

Step 1.2.5.2.2

Divide each term in $2.23606797 x = - 1$ by $2.23606797$ and simplify.

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Step 1.2.5.2.2.1

Divide each term in $2.23606797 x = - 1$ by $2.23606797$ .

$\frac{2.23606797 x}{2.23606797} = \frac{- 1}{2.23606797}$

Step 1.2.5.2.2.2

Simplify the left side.

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Step 1.2.5.2.2.2.1

Cancel the common factor of $2.23606797$ .

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Step 1.2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2.23606797 x}{2.23606797} = \frac{- 1}{2.23606797}$

Step 1.2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{2.23606797}$

Step 1.2.5.2.2.3

Simplify the right side.

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Step 1.2.5.2.2.3.1

Divide $- 1$ by $2.23606797$ .

$x = - 0.44721359$

Step 1.2.6

Set $2.23606797 x - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.6.1

Set $2.23606797 x - 1$ equal to $0$ .

$2.23606797 x - 1 = 0$

Step 1.2.6.2

Solve $2.23606797 x - 1 = 0$ for $x$ .

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Step 1.2.6.2.1

Add $1$ to both sides of the equation.

$2.23606797 x = 1$

Step 1.2.6.2.2

Divide each term in $2.23606797 x = 1$ by $2.23606797$ and simplify.

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Step 1.2.6.2.2.1

Divide each term in $2.23606797 x = 1$ by $2.23606797$ .

$\frac{2.23606797 x}{2.23606797} = \frac{1}{2.23606797}$

Step 1.2.6.2.2.2

Simplify the left side.

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Step 1.2.6.2.2.2.1

Cancel the common factor of $2.23606797$ .

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Step 1.2.6.2.2.2.1.1

Cancel the common factor.

$\frac{2.23606797 x}{2.23606797} = \frac{1}{2.23606797}$

Step 1.2.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2.23606797}$

Step 1.2.6.2.2.3

Simplify the right side.

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Step 1.2.6.2.2.3.1

Divide $1$ by $2.23606797$ .

$x = 0.44721359$

Step 1.2.7

The final solution is all the values that make $5 e^{- 2.5 x^{2}} (2.23606797 x + 1) (2.23606797 x - 1) = 0$ true.

$x = - 0.44721359, 0.44721359$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - 0.44721359) \cup (- 0.44721359, 0.44721359) \cup (0.44721359, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - 0.44721359)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 3$ in the expression.

$f'' (- 3) = 25 {(- 3)}^{2} e^{- 2.5 {(- 3)}^{2}} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $- 3$ to the power of $2$ .

$f'' (- 3) = 25 \cdot (9 e^{- 2.5 {(- 3)}^{2}}) - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.2

Multiply $25$ by $9$ .

$f'' (- 3) = 225 e^{- 2.5 {(- 3)}^{2}} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.3

Raise $- 3$ to the power of $2$ .

$f'' (- 3) = 225 e^{- 2.5 \cdot 9} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.4

Multiply $- 2.5$ by $9$ .

$f'' (- 3) = 225 e^{- 22.5} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.5

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

$f'' (- 3) = 225 (\frac{1}{e^{22.5}}) - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.6

Combine $225$ and $\frac{1}{e^{22.5}}$ .

$f'' (- 3) = \frac{225}{e^{22.5}} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.7

Replace $e$ with an approximation.

$f'' (- 3) = \frac{225}{{2.71828182}^{22.5}} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.8

Raise $2.71828182$ to the power of $22.5$ .

$f'' (- 3) = \frac{225}{5910522063.02304} - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.9

Divide $225$ by $5910522063.02304$ .

$f'' (- 3) = 0.00000003 - 5 e^{- 2.5 {(- 3)}^{2}}$

Step 4.2.1.10

Raise $- 3$ to the power of $2$ .

$f'' (- 3) = 0.00000003 - 5 e^{- 2.5 \cdot 9}$

Step 4.2.1.11

Multiply $- 2.5$ by $9$ .

$f'' (- 3) = 0.00000003 - 5 e^{- 22.5}$

Step 4.2.1.12

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

$f'' (- 3) = 0.00000003 - 5 \frac{1}{e^{22.5}}$

Step 4.2.1.13

Combine $- 5$ and $\frac{1}{e^{22.5}}$ .

$f'' (- 3) = 0.00000003 + \frac{- 5}{e^{22.5}}$

Step 4.2.1.14

Move the negative in front of the fraction.

$f'' (- 3) = 0.00000003 - \frac{5}{e^{22.5}}$

Step 4.2.1.15

Replace $e$ with an approximation.

$f'' (- 3) = 0.00000003 - \frac{5}{{2.71828182}^{22.5}}$

Step 4.2.1.16

Raise $2.71828182$ to the power of $22.5$ .

$f'' (- 3) = 0.00000003 - \frac{5}{5910522063.02304}$

Step 4.2.1.17

Divide $5$ by $5910522063.02304$ .

$f'' (- 3) = 0.00000003 - 1 \cdot 0$

Step 4.2.1.18

Multiply $- 1$ by $0$ .

$f'' (- 3) = 0.00000003 0$

Step 4.2.2

Subtract $0$ from $0.00000003$ .

$f'' (- 3) = 0.00000003$

Step 4.2.3

The final answer is $0.00000003$ .

$0.00000003$

Step 4.3

The graph is concave up on the interval $(- \infty, - 0.44721359)$ because $f'' (- 3)$ is positive.

Concave up on $(- \infty, - 0.44721359)$ since $f'' (x)$ is positive

Step 5

Substitute any number from the interval $(- 0.44721359, 0.44721359)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = 25 {(0)}^{2} e^{- 2.5 {(0)}^{2}} - 5 e^{- 2.5 {(0)}^{2}}$

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

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

$f'' (0) = 25 \cdot (0 e^{- 2.5 {(0)}^{2}}) - 5 e^{- 2.5 {(0)}^{2}}$

Step 5.2.1.2

Multiply $25$ by $0$ .

$f'' (0) = 0 e^{- 2.5 {(0)}^{2}} - 5 e^{- 2.5 {(0)}^{2}}$

Step 5.2.1.3

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

$f'' (0) = 0 e^{- 2.5 \cdot 0} - 5 e^{- 2.5 {(0)}^{2}}$

Step 5.2.1.4

Multiply $- 2.5$ by $0$ .

$f'' (0) = 0 e^{0} - 5 e^{- 2.5 {(0)}^{2}}$

Step 5.2.1.5

Anything raised to $0$ is $1$ .

$f'' (0) = 0 \cdot 1 - 5 e^{- 2.5 {(0)}^{2}}$

Step 5.2.1.6

Multiply $0$ by $1$ .

$f'' (0) = 0 - 5 e^{- 2.5 {(0)}^{2}}$

Step 5.2.1.7

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

$f'' (0) = 0 - 5 e^{- 2.5 \cdot 0}$

Step 5.2.1.8

Multiply $- 2.5$ by $0$ .

$f'' (0) = 0 - 5 e^{0}$

Step 5.2.1.9

Anything raised to $0$ is $1$ .

$f'' (0) = 0 - 5 \cdot 1$

Step 5.2.1.10

Multiply $- 5$ by $1$ .

$f'' (0) = 0 - 5$

Step 5.2.2

Subtract $5$ from $0$ .

$f'' (0) = - 5$

Step 5.2.3

The final answer is $- 5$ .

$- 5$

Step 5.3

The graph is concave down on the interval $(- 0.44721359, 0.44721359)$ because $f'' (0)$ is negative.

Concave down on $(- 0.44721359, 0.44721359)$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(0.44721359, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $3$ in the expression.

$f'' (3) = 25 {(3)}^{2} e^{- 2.5 {(3)}^{2}} - 5 e^{- 2.5 {(3)}^{2}}$

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

Raise $3$ to the power of $2$ .

$f'' (3) = 25 \cdot (9 e^{- 2.5 {(3)}^{2}}) - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.2

Multiply $25$ by $9$ .

$f'' (3) = 225 e^{- 2.5 {(3)}^{2}} - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.3

Raise $3$ to the power of $2$ .

$f'' (3) = 225 e^{- 2.5 \cdot 9} - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.4

Multiply $- 2.5$ by $9$ .

$f'' (3) = 225 e^{- 22.5} - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.5

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

$f'' (3) = 225 (\frac{1}{e^{22.5}}) - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.6

Combine $225$ and $\frac{1}{e^{22.5}}$ .

$f'' (3) = \frac{225}{e^{22.5}} - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.7

Replace $e$ with an approximation.

$f'' (3) = \frac{225}{{2.71828182}^{22.5}} - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.8

Raise $2.71828182$ to the power of $22.5$ .

$f'' (3) = \frac{225}{5910522063.02304} - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.9

Divide $225$ by $5910522063.02304$ .

$f'' (3) = 0.00000003 - 5 e^{- 2.5 {(3)}^{2}}$

Step 6.2.1.10

Raise $3$ to the power of $2$ .

$f'' (3) = 0.00000003 - 5 e^{- 2.5 \cdot 9}$

Step 6.2.1.11

Multiply $- 2.5$ by $9$ .

$f'' (3) = 0.00000003 - 5 e^{- 22.5}$

Step 6.2.1.12

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

$f'' (3) = 0.00000003 - 5 \frac{1}{e^{22.5}}$

Step 6.2.1.13

Combine $- 5$ and $\frac{1}{e^{22.5}}$ .

$f'' (3) = 0.00000003 + \frac{- 5}{e^{22.5}}$

Step 6.2.1.14

Move the negative in front of the fraction.

$f'' (3) = 0.00000003 - \frac{5}{e^{22.5}}$

Step 6.2.1.15

Replace $e$ with an approximation.

$f'' (3) = 0.00000003 - \frac{5}{{2.71828182}^{22.5}}$

Step 6.2.1.16

Raise $2.71828182$ to the power of $22.5$ .

$f'' (3) = 0.00000003 - \frac{5}{5910522063.02304}$

Step 6.2.1.17

Divide $5$ by $5910522063.02304$ .

$f'' (3) = 0.00000003 - 1 \cdot 0$

Step 6.2.1.18

Multiply $- 1$ by $0$ .

$f'' (3) = 0.00000003 0$

Step 6.2.2

Subtract $0$ from $0.00000003$ .

$f'' (3) = 0.00000003$

Step 6.2.3

The final answer is $0.00000003$ .

$0.00000003$

Step 6.3

The graph is concave up on the interval $(0.44721359, \infty)$ because $f'' (3)$ is positive.

Concave up on $(0.44721359, \infty)$ since $f'' (x)$ is positive

Step 7

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $(- \infty, - 0.44721359)$ since $f'' (x)$ is positive

Concave down on $(- 0.44721359, 0.44721359)$ since $f'' (x)$ is negative

Concave up on $(0.44721359, \infty)$ since $f'' (x)$ is positive

Step 8