Calculus Examples

$e^{4 x} + e^{- x}$

Step 1

Write $e^{4 x} + e^{- x}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

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

$\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [e^{- x}]$

Step 2.2

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

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Step 2.2.1

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 $g (x) = 4 x$ .

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Step 2.2.1.1

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

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.2.1.2

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

$e^{u_{1}} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.2.1.3

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

$e^{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.2.2

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

$e^{4 x} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [e^{- x}]$

Step 2.2.3

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

$e^{4 x} (4 \cdot 1) + \frac{d}{d x} [e^{- x}]$

Step 2.2.4

Multiply $4$ by $1$ .

$e^{4 x} \cdot 4 + \frac{d}{d x} [e^{- x}]$

Step 2.2.5

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

$4 e^{4 x} + \frac{d}{d x} [e^{- x}]$

Step 2.3

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

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Step 2.3.1

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 $g (x) = - x$ .

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Step 2.3.1.1

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

$4 e^{4 x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x]$

Step 2.3.1.2

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

$4 e^{4 x} + e^{u_{2}} \frac{d}{d x} [- x]$

Step 2.3.1.3

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

$4 e^{4 x} + e^{- x} \frac{d}{d x} [- x]$

Step 2.3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$4 e^{4 x} + e^{- x} (- \frac{d}{d x} [x])$

Step 2.3.3

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

$4 e^{4 x} + e^{- x} (- 1 \cdot 1)$

Step 2.3.4

Multiply $- 1$ by $1$ .

$4 e^{4 x} + e^{- x} \cdot - 1$

Step 2.3.5

Move $- 1$ to the left of $e^{- x}$ .

$4 e^{4 x} - 1 \cdot e^{- x}$

Step 2.3.6

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$4 e^{4 x} - e^{- x}$

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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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Step 3.2.2.1

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

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

Step 3.2.2.2

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

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

Step 3.2.2.3

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

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

Multiply $4$ by $1$ .

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

Step 3.2.6

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

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

Step 3.2.7

Multiply $4$ by $4$ .

$f'' (x) = 16 e^{4 x} + \frac{d}{d x} (- e^{- x})$

Step 3.3

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

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Step 3.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{- x}$ with respect to $x$ is $- \frac{d}{d x} [e^{- x}]$ .

$f'' (x) = 16 e^{4 x} - \frac{d}{d x} e^{- x}$

Step 3.3.2

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 $g (x) = - x$ .

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Step 3.3.2.1

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

$f'' (x) = 16 e^{4 x} - (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x))$

Step 3.3.2.2

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

$f'' (x) = 16 e^{4 x} - (e^{u_{2}} \frac{d}{d x} (- x))$

Step 3.3.2.3

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

$f'' (x) = 16 e^{4 x} - (e^{- x} \frac{d}{d x} (- x))$

Step 3.3.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$f'' (x) = 16 e^{4 x} - (e^{- x} (- \frac{d}{d x} x))$

Step 3.3.4

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

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

Step 3.3.5

Multiply $- 1$ by $1$ .

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

Step 3.3.6

Move $- 1$ to the left of $e^{- x}$ .

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

Step 3.3.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 3.3.8

Multiply $- 1$ by $- 1$ .

$f'' (x) = 16 e^{4 x} + 1 e^{- x}$

Step 3.3.9

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

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

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$4 e^{4 x} - e^{- x} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

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

$\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [e^{- x}]$

Step 5.1.2

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

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Step 5.1.2.1

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 $g (x) = 4 x$ .

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Step 5.1.2.1.1

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

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 5.1.2.1.2

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

$e^{u_{1}} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 5.1.2.1.3

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

$e^{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 5.1.2.2

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

$e^{4 x} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [e^{- x}]$

Step 5.1.2.3

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

$e^{4 x} (4 \cdot 1) + \frac{d}{d x} [e^{- x}]$

Step 5.1.2.4

Multiply $4$ by $1$ .

$e^{4 x} \cdot 4 + \frac{d}{d x} [e^{- x}]$

Step 5.1.2.5

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

$4 e^{4 x} + \frac{d}{d x} [e^{- x}]$

Step 5.1.3

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

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Step 5.1.3.1

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 $g (x) = - x$ .

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Step 5.1.3.1.1

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

$4 e^{4 x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x]$

Step 5.1.3.1.2

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

$4 e^{4 x} + e^{u_{2}} \frac{d}{d x} [- x]$

Step 5.1.3.1.3

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

$4 e^{4 x} + e^{- x} \frac{d}{d x} [- x]$

Step 5.1.3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$4 e^{4 x} + e^{- x} (- \frac{d}{d x} [x])$

Step 5.1.3.3

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

$4 e^{4 x} + e^{- x} (- 1 \cdot 1)$

Step 5.1.3.4

Multiply $- 1$ by $1$ .

$4 e^{4 x} + e^{- x} \cdot - 1$

Step 5.1.3.5

Move $- 1$ to the left of $e^{- x}$ .

$4 e^{4 x} - 1 \cdot e^{- x}$

Step 5.1.3.6

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $4 e^{4 x} - e^{- x}$ .

$4 e^{4 x} - e^{- x}$

Step 6

Set the first derivative equal to $0$ then solve the equation $4 e^{4 x} - e^{- x} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$4 e^{4 x} - e^{- x} = 0$

Step 6.2

Move $- e^{- x}$ to the right side of the equation by adding it to both sides.

$4 e^{4 x} = e^{- x}$

Step 6.3

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

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

Step 6.4

Expand the left side.

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Step 6.4.1

Rewrite $\ln (4 e^{4 x})$ as $\ln (4) + \ln (e^{4 x})$ .

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

Step 6.4.2

Expand $\ln (e^{4 x})$ by moving $4 x$ outside the logarithm.

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

Step 6.4.3

The natural logarithm of $e$ is $1$ .

$\ln (4) + 4 x \cdot 1 = \ln (e^{- x})$

Step 6.4.4

Multiply $4$ by $1$ .

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

Step 6.5

Expand the right side.

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Step 6.5.1

Expand $\ln (e^{- x})$ by moving $- x$ outside the logarithm.

$\ln (4) + 4 x = - x \ln (e)$

Step 6.5.2

The natural logarithm of $e$ is $1$ .

$\ln (4) + 4 x = - x \cdot 1$

Step 6.5.3

Multiply $- 1$ by $1$ .

$\ln (4) + 4 x = - x$

Step 6.6

Move all terms containing $x$ to the left side of the equation.

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Step 6.6.1

Add $x$ to both sides of the equation.

$\ln (4) + 4 x + x = 0$

Step 6.6.2

Add $4 x$ and $x$ .

$\ln (4) + 5 x = 0$

Step 6.7

Subtract $\ln (4)$ from both sides of the equation.

$5 x = - \ln (4)$

Step 6.8

Divide each term in $5 x = - \ln (4)$ by $5$ and simplify.

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Step 6.8.1

Divide each term in $5 x = - \ln (4)$ by $5$ .

$\frac{5 x}{5} = \frac{- \ln (4)}{5}$

Step 6.8.2

Simplify the left side.

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Step 6.8.2.1

Cancel the common factor of $5$ .

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Step 6.8.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{- \ln (4)}{5}$

Step 6.8.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (4)}{5}$

Step 6.8.3

Simplify the right side.

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Step 6.8.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (4)}{5}$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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.

Step 8

Critical points to evaluate.

$x = - \frac{\ln (4)}{5}$

Step 9

Evaluate the second derivative at $x = - \frac{\ln (4)}{5}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$16 e^{4 (- \frac{\ln (4)}{5})} + e^{- (- \frac{\ln (4)}{5})}$

Step 10

Simplify each term.

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Step 10.1

Rewrite $\frac{\ln (4)}{5}$ as $\frac{1}{5} \ln (4)$ .

$16 e^{4 (- (\frac{1}{5} \ln (4)))} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.2

Simplify $\frac{1}{5} \ln (4)$ by moving $\frac{1}{5}$ inside the logarithm.

$16 e^{4 (- \ln (4^{\frac{1}{5}}))} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.3

Multiply $4 (- \ln (4^{\frac{1}{5}}))$ .

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Step 10.3.1

Multiply $- 1$ by $4$ .

$16 e^{- 4 \ln (4^{\frac{1}{5}})} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.3.2

Simplify $- 4 \ln (4^{\frac{1}{5}})$ by moving $4$ inside the logarithm.

$16 e^{- \ln ({(4^{\frac{1}{5}})}^{4})} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.4

Simplify $- \ln ({(4^{\frac{1}{5}})}^{4})$ by moving $- 1$ inside the logarithm.

$16 e^{\ln ({({(4^{\frac{1}{5}})}^{4})}^{- 1})} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.5

Exponentiation and log are inverse functions.

$16 {({(4^{\frac{1}{5}})}^{4})}^{- 1} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.6

Multiply the exponents in ${({(4^{\frac{1}{5}})}^{4})}^{- 1}$ .

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Step 10.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 {(4^{\frac{1}{5}})}^{4 \cdot - 1} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.6.2

Multiply $4$ by $- 1$ .

$16 {(4^{\frac{1}{5}})}^{- 4} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.7

Multiply the exponents in ${(4^{\frac{1}{5}})}^{- 4}$ .

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Step 10.7.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 \cdot 4^{\frac{1}{5} \cdot - 4} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.7.2

Combine $\frac{1}{5}$ and $- 4$ .

$16 \cdot 4^{\frac{- 4}{5}} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.7.3

Move the negative in front of the fraction.

$16 \cdot 4^{- \frac{4}{5}} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.8

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

$16 \cdot \frac{1}{4^{\frac{4}{5}}} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.9

Combine $16$ and $\frac{1}{4^{\frac{4}{5}}}$ .

$\frac{16}{4^{\frac{4}{5}}} + e^{- (- \frac{\ln (4)}{5})}$

Step 10.10

Rewrite $\frac{\ln (4)}{5}$ as $\frac{1}{5} \ln (4)$ .

$\frac{16}{4^{\frac{4}{5}}} + e^{- - (\frac{1}{5} \ln (4))}$

Step 10.11

Simplify $\frac{1}{5} \ln (4)$ by moving $\frac{1}{5}$ inside the logarithm.

$\frac{16}{4^{\frac{4}{5}}} + e^{- - \ln (4^{\frac{1}{5}})}$

Step 10.12

Multiply $- - \ln (4^{\frac{1}{5}})$ .

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Step 10.12.1

Multiply $- 1$ by $- 1$ .

$\frac{16}{4^{\frac{4}{5}}} + e^{1 \ln (4^{\frac{1}{5}})}$

Step 10.12.2

Multiply $\ln (4^{\frac{1}{5}})$ by $1$ .

$\frac{16}{4^{\frac{4}{5}}} + e^{\ln (4^{\frac{1}{5}})}$

Step 10.13

Exponentiation and log are inverse functions.

$\frac{16}{4^{\frac{4}{5}}} + 4^{\frac{1}{5}}$

Step 11

$x = - \frac{\ln (4)}{5}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - \frac{\ln (4)}{5}$ is a local minimum

Step 12

Find the y-value when $x = - \frac{\ln (4)}{5}$ .

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Step 12.1

Simplify $- \frac{\ln (4)}{5}$ to substitute in $f (x) = e^{4 x} + e^{- x}$ .

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Step 12.1.1

Rewrite $\frac{\ln (4)}{5}$ as $\frac{1}{5} \ln (4)$ .

$y = - (\frac{1}{5} \cdot \ln (4))$

Step 12.1.2

Simplify $\frac{1}{5} \ln (4)$ by moving $\frac{1}{5}$ inside the logarithm.

$y = - \ln (4^{\frac{1}{5}})$

Step 12.2

Replace the variable $x$ with $- \ln (4^{\frac{1}{5}})$ in the expression.

$f (- \ln (4^{\frac{1}{5}})) = e^{4 (- \ln (4^{\frac{1}{5}}))} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3

Simplify the result.

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Step 12.3.1

Simplify each term.

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Step 12.3.1.1

Multiply $4 (- \ln (4^{\frac{1}{5}}))$ .

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Step 12.3.1.1.1

Multiply $- 1$ by $4$ .

$f (- \ln (4^{\frac{1}{5}})) = e^{- 4 \ln (4^{\frac{1}{5}})} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.1.2

Simplify $- 4 \ln (4^{\frac{1}{5}})$ by moving $4$ inside the logarithm.

$f (- \ln (4^{\frac{1}{5}})) = e^{- \ln ({(4^{\frac{1}{5}})}^{4})} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.2

Simplify $- \ln ({(4^{\frac{1}{5}})}^{4})$ by moving $- 1$ inside the logarithm.

$f (- \ln (4^{\frac{1}{5}})) = e^{\ln ({({(4^{\frac{1}{5}})}^{4})}^{- 1})} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.3

Exponentiation and log are inverse functions.

$f (- \ln (4^{\frac{1}{5}})) = {({(4^{\frac{1}{5}})}^{4})}^{- 1} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.4

Multiply the exponents in ${({(4^{\frac{1}{5}})}^{4})}^{- 1}$ .

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Step 12.3.1.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \ln (4^{\frac{1}{5}})) = {(4^{\frac{1}{5}})}^{4 \cdot - 1} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.4.2

Multiply $4$ by $- 1$ .

$f (- \ln (4^{\frac{1}{5}})) = {(4^{\frac{1}{5}})}^{- 4} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.5

Multiply the exponents in ${(4^{\frac{1}{5}})}^{- 4}$ .

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Step 12.3.1.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \ln (4^{\frac{1}{5}})) = 4^{\frac{1}{5} \cdot - 4} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.5.2

Combine $\frac{1}{5}$ and $- 4$ .

$f (- \ln (4^{\frac{1}{5}})) = 4^{\frac{- 4}{5}} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.5.3

Move the negative in front of the fraction.

$f (- \ln (4^{\frac{1}{5}})) = 4^{- \frac{4}{5}} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.6

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

$f (- \ln (4^{\frac{1}{5}})) = \frac{1}{4^{\frac{4}{5}}} + e^{- (- \ln (4^{\frac{1}{5}}))}$

Step 12.3.1.7

Multiply $- (- \ln (4^{\frac{1}{5}}))$ .

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Step 12.3.1.7.1

Multiply $- 1$ by $- 1$ .

$f (- \ln (4^{\frac{1}{5}})) = \frac{1}{4^{\frac{4}{5}}} + e^{1 \ln (4^{\frac{1}{5}})}$

Step 12.3.1.7.2

Multiply $\ln (4^{\frac{1}{5}})$ by $1$ .

$f (- \ln (4^{\frac{1}{5}})) = \frac{1}{4^{\frac{4}{5}}} + e^{\ln (4^{\frac{1}{5}})}$

Step 12.3.1.8

Exponentiation and log are inverse functions.

$f (- \ln (4^{\frac{1}{5}})) = \frac{1}{4^{\frac{4}{5}}} + 4^{\frac{1}{5}}$

Step 12.3.2

The final answer is $\frac{1}{4^{\frac{4}{5}}} + 4^{\frac{1}{5}}$ .

$y = \frac{1}{4^{\frac{4}{5}}} + 4^{\frac{1}{5}}$

Step 13

These are the local extrema for $f (x) = e^{4 x} + e^{- x}$ .

$(- \frac{\ln (4)}{5}, \frac{1}{4^{\frac{4}{5}}} + 4^{\frac{1}{5}})$ is a local minima

Step 14