Calculus Examples

$f (x) = e^{3 x} + e^{- x}$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

By the Sum Rule, the derivative of $e^{3 x} + e^{- x}$ with respect to $x$ is $\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [e^{- x}]$ .

$\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [e^{- x}]$

Step 1.2

Evaluate $\frac{d}{d x} [e^{3 x}]$ .

Tap for more steps...

Step 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 $g (x) = 3 x$ .

Tap for more steps...

Step 1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $3 x$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [3 x] + \frac{d}{d x} [e^{- x}]$

Step 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 $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [3 x] + \frac{d}{d x} [e^{- x}]$

Step 1.2.1.3

Replace all occurrences of $u_{1}$ with $3 x$ .

$e^{3 x} \frac{d}{d x} [3 x] + \frac{d}{d x} [e^{- x}]$

Step 1.2.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$e^{3 x} (3 \frac{d}{d x} [x]) + \frac{d}{d x} [e^{- x}]$

Step 1.2.3

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

$e^{3 x} (3 \cdot 1) + \frac{d}{d x} [e^{- x}]$

Step 1.2.4

Multiply $3$ by $1$ .

$e^{3 x} \cdot 3 + \frac{d}{d x} [e^{- x}]$

Step 1.2.5

Move $3$ to the left of $e^{3 x}$ .

$3 e^{3 x} + \frac{d}{d x} [e^{- x}]$

Step 1.3

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

Tap for more steps...

Step 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 $g (x) = - x$ .

Tap for more steps...

Step 1.3.1.1

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

$3 e^{3 x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x]$

Step 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 $a$ = $e$ .

$3 e^{3 x} + e^{u_{2}} \frac{d}{d x} [- x]$

Step 1.3.1.3

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

$3 e^{3 x} + e^{- x} \frac{d}{d x} [- x]$

Step 1.3.2

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

$3 e^{3 x} + e^{- x} (- \frac{d}{d x} [x])$

Step 1.3.3

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

$3 e^{3 x} + e^{- x} (- 1 \cdot 1)$

Step 1.3.4

Multiply $- 1$ by $1$ .

$3 e^{3 x} + e^{- x} \cdot - 1$

Step 1.3.5

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

$3 e^{3 x} - 1 \cdot e^{- x}$

Step 1.3.6

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

$3 e^{3 x} - e^{- x}$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $3 e^{3 x} - e^{- x}$ with respect to $x$ is $\frac{d}{d x} [3 e^{3 x}] + \frac{d}{d x} [- e^{- x}]$ .

$f'' (x) = \frac{d}{d x} (3 e^{3 x}) + \frac{d}{d x} (- e^{- x})$

Step 2.2

Evaluate $\frac{d}{d x} [3 e^{3 x}]$ .

Tap for more steps...

Step 2.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{3 x}$ with respect to $x$ is $3 \frac{d}{d x} [e^{3 x}]$ .

$f'' (x) = 3 \frac{d}{d x} (e^{3 x}) + \frac{d}{d x} (- e^{- x})$

Step 2.2.2

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 $g (x) = 3 x$ .

Tap for more steps...

Step 2.2.2.1

To apply the Chain Rule, set $u_{1}$ as $3 x$ .

$f'' (x) = 3 (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (3 x)) + \frac{d}{d x} (- e^{- x})$

Step 2.2.2.2

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

$f'' (x) = 3 (e^{u_{1}} \frac{d}{d x} (3 x)) + \frac{d}{d x} (- e^{- x})$

Step 2.2.2.3

Replace all occurrences of $u_{1}$ with $3 x$ .

$f'' (x) = 3 (e^{3 x} \frac{d}{d x} (3 x)) + \frac{d}{d x} (- e^{- x})$

Step 2.2.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$f'' (x) = 3 (e^{3 x} (3 \frac{d}{d x} (x))) + \frac{d}{d x} (- e^{- x})$

Step 2.2.4

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

$f'' (x) = 3 (e^{3 x} (3 \cdot 1)) + \frac{d}{d x} (- e^{- x})$

Step 2.2.5

Multiply $3$ by $1$ .

$f'' (x) = 3 (e^{3 x} \cdot 3) + \frac{d}{d x} (- e^{- x})$

Step 2.2.6

Move $3$ to the left of $e^{3 x}$ .

$f'' (x) = 3 (3 \cdot e^{3 x}) + \frac{d}{d x} (- e^{- x})$

Step 2.2.7

Multiply $3$ by $3$ .

$f'' (x) = 9 e^{3 x} + \frac{d}{d x} (- e^{- x})$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$f'' (x) = 9 e^{3 x} - \frac{d}{d x} e^{- x}$

Step 2.3.2

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 $g (x) = - x$ .

Tap for more steps...

Step 2.3.2.1

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

$f'' (x) = 9 e^{3 x} - (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x))$

Step 2.3.2.2

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

$f'' (x) = 9 e^{3 x} - (e^{u_{2}} \frac{d}{d x} (- x))$

Step 2.3.2.3

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

$f'' (x) = 9 e^{3 x} - (e^{- x} \frac{d}{d x} (- x))$

Step 2.3.3

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

$f'' (x) = 9 e^{3 x} - (e^{- x} (- \frac{d}{d x} x))$

Step 2.3.4

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

$f'' (x) = 9 e^{3 x} - (e^{- x} (- 1 \cdot 1))$

Step 2.3.5

Multiply $- 1$ by $1$ .

$f'' (x) = 9 e^{3 x} - (e^{- x} \cdot - 1)$

Step 2.3.6

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

$f'' (x) = 9 e^{3 x} - (- 1 \cdot e^{- x})$

Step 2.3.7

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

$f'' (x) = 9 e^{3 x} + e^{- x}$

Step 2.3.8

Multiply $- 1$ by $- 1$ .

$f'' (x) = 9 e^{3 x} + 1 e^{- x}$

Step 2.3.9

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

$f'' (x) = 9 e^{3 x} + e^{- x}$

Step 3

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

$3 e^{3 x} - e^{- x} = 0$

Step 4

Find the first derivative.

Tap for more steps...

Step 4.1

Find the first derivative.

Tap for more steps...

Step 4.1.1

By the Sum Rule, the derivative of $e^{3 x} + e^{- x}$ with respect to $x$ is $\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [e^{- x}]$ .

$f' (x) = \frac{d}{d x} (e^{3 x}) + \frac{d}{d x} (e^{- x})$

Step 4.1.2

Evaluate $\frac{d}{d x} [e^{3 x}]$ .

Tap for more steps...

Step 4.1.2.1

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 $g (x) = 3 x$ .

Tap for more steps...

Step 4.1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $3 x$ .

$f' (x) = \frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (3 x) + \frac{d}{d x} (e^{- x})$

Step 4.1.2.1.2

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

$f' (x) = e^{u_{1}} \frac{d}{d x} (3 x) + \frac{d}{d x} (e^{- x})$

Step 4.1.2.1.3

Replace all occurrences of $u_{1}$ with $3 x$ .

$f' (x) = e^{3 x} \frac{d}{d x} (3 x) + \frac{d}{d x} (e^{- x})$

Step 4.1.2.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$f' (x) = e^{3 x} (3 \frac{d}{d x} (x)) + \frac{d}{d x} (e^{- x})$

Step 4.1.2.3

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

$f' (x) = e^{3 x} (3 \cdot 1) + \frac{d}{d x} (e^{- x})$

Step 4.1.2.4

Multiply $3$ by $1$ .

$f' (x) = e^{3 x} \cdot 3 + \frac{d}{d x} (e^{- x})$

Step 4.1.2.5

Move $3$ to the left of $e^{3 x}$ .

$f' (x) = 3 e^{3 x} + \frac{d}{d x} (e^{- x})$

Step 4.1.3

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

Tap for more steps...

Step 4.1.3.1

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 $g (x) = - x$ .

Tap for more steps...

Step 4.1.3.1.1

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

$f' (x) = 3 e^{3 x} + \frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x)$

Step 4.1.3.1.2

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

$f' (x) = 3 e^{3 x} + e^{u_{2}} \frac{d}{d x} (- x)$

Step 4.1.3.1.3

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

$f' (x) = 3 e^{3 x} + e^{- x} \frac{d}{d x} (- x)$

Step 4.1.3.2

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

$f' (x) = 3 e^{3 x} + e^{- x} (- \frac{d}{d x} x)$

Step 4.1.3.3

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

$f' (x) = 3 e^{3 x} + e^{- x} (- 1 \cdot 1)$

Step 4.1.3.4

Multiply $- 1$ by $1$ .

$f' (x) = 3 e^{3 x} + e^{- x} \cdot - 1$

Step 4.1.3.5

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

$f' (x) = 3 e^{3 x} - 1 \cdot e^{- x}$

Step 4.1.3.6

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

$f' (x) = 3 e^{3 x} - e^{- x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $3 e^{3 x} - e^{- x}$ .

$3 e^{3 x} - e^{- x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $3 e^{3 x} - e^{- x} = 0$ .

Tap for more steps...

Step 5.1

Set the first derivative equal to $0$ .

$3 e^{3 x} - e^{- x} = 0$

Step 5.2

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

$3 e^{3 x} = e^{- x}$

Step 5.3

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

$\ln (3 e^{3 x}) = \ln (e^{- x})$

Step 5.4

Expand the left side.

Tap for more steps...

Step 5.4.1

Rewrite $\ln (3 e^{3 x})$ as $\ln (3) + \ln (e^{3 x})$ .

$\ln (3) + \ln (e^{3 x}) = \ln (e^{- x})$

Step 5.4.2

Expand $\ln (e^{3 x})$ by moving $3 x$ outside the logarithm.

$\ln (3) + 3 x \ln (e) = \ln (e^{- x})$

Step 5.4.3

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

$\ln (3) + 3 x \cdot 1 = \ln (e^{- x})$

Step 5.4.4

Multiply $3$ by $1$ .

$\ln (3) + 3 x = \ln (e^{- x})$

Step 5.5

Expand the right side.

Tap for more steps...

Step 5.5.1

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

$\ln (3) + 3 x = - x \ln (e)$

Step 5.5.2

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

$\ln (3) + 3 x = - x \cdot 1$

Step 5.5.3

Multiply $- 1$ by $1$ .

$\ln (3) + 3 x = - x$

Step 5.6

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

Tap for more steps...

Step 5.6.1

Add $x$ to both sides of the equation.

$\ln (3) + 3 x + x = 0$

Step 5.6.2

Add $3 x$ and $x$ .

$\ln (3) + 4 x = 0$

Step 5.7

Subtract $\ln (3)$ from both sides of the equation.

$4 x = - \ln (3)$

Step 5.8

Divide each term in $4 x = - \ln (3)$ by $4$ and simplify.

Tap for more steps...

Step 5.8.1

Divide each term in $4 x = - \ln (3)$ by $4$ .

$\frac{4 x}{4} = \frac{- \ln (3)}{4}$

Step 5.8.2

Simplify the left side.

Tap for more steps...

Step 5.8.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 5.8.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{- \ln (3)}{4}$

Step 5.8.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (3)}{4}$

Step 5.8.3

Simplify the right side.

Tap for more steps...

Step 5.8.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (3)}{4}$

Step 6

Find the values where the derivative is undefined.

Tap for more steps...

Step 6.1

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 7

Critical points to evaluate.

$x = - \frac{\ln (3)}{4}$

Step 8

Evaluate the second derivative at $x = - \frac{\ln (3)}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$9 e^{3 (- \frac{\ln (3)}{4})} + e^{- (- \frac{\ln (3)}{4})}$

Step 9

Simplify each term.

Tap for more steps...

Step 9.1

Rewrite $\frac{\ln (3)}{4}$ as $\frac{1}{4} \ln (3)$ .

$9 e^{3 (- (\frac{1}{4} \ln (3)))} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.2

Simplify $\frac{1}{4} \ln (3)$ by moving $\frac{1}{4}$ inside the logarithm.

$9 e^{3 (- \ln (3^{\frac{1}{4}}))} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.3

Multiply $3 (- \ln (3^{\frac{1}{4}}))$ .

Tap for more steps...

Step 9.3.1

Multiply $- 1$ by $3$ .

$9 e^{- 3 \ln (3^{\frac{1}{4}})} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.3.2

Simplify $- 3 \ln (3^{\frac{1}{4}})$ by moving $3$ inside the logarithm.

$9 e^{- \ln ({(3^{\frac{1}{4}})}^{3})} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.4

Simplify $- \ln ({(3^{\frac{1}{4}})}^{3})$ by moving $- 1$ inside the logarithm.

$9 e^{\ln ({({(3^{\frac{1}{4}})}^{3})}^{- 1})} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.5

Exponentiation and log are inverse functions.

$9 {({(3^{\frac{1}{4}})}^{3})}^{- 1} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.6

Multiply the exponents in ${({(3^{\frac{1}{4}})}^{3})}^{- 1}$ .

Tap for more steps...

Step 9.6.1

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

$9 {(3^{\frac{1}{4}})}^{3 \cdot - 1} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.6.2

Multiply $3$ by $- 1$ .

$9 {(3^{\frac{1}{4}})}^{- 3} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.7

Multiply the exponents in ${(3^{\frac{1}{4}})}^{- 3}$ .

Tap for more steps...

Step 9.7.1

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

$9 \cdot 3^{\frac{1}{4} \cdot - 3} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.7.2

Combine $\frac{1}{4}$ and $- 3$ .

$9 \cdot 3^{\frac{- 3}{4}} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.7.3

Move the negative in front of the fraction.

$9 \cdot 3^{- \frac{3}{4}} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.8

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

$9 \cdot \frac{1}{3^{\frac{3}{4}}} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.9

Combine $9$ and $\frac{1}{3^{\frac{3}{4}}}$ .

$\frac{9}{3^{\frac{3}{4}}} + e^{- (- \frac{\ln (3)}{4})}$

Step 9.10

Rewrite $\frac{\ln (3)}{4}$ as $\frac{1}{4} \ln (3)$ .

$\frac{9}{3^{\frac{3}{4}}} + e^{- - (\frac{1}{4} \ln (3))}$

Step 9.11

Simplify $\frac{1}{4} \ln (3)$ by moving $\frac{1}{4}$ inside the logarithm.

$\frac{9}{3^{\frac{3}{4}}} + e^{- - \ln (3^{\frac{1}{4}})}$

Step 9.12

Multiply $- - \ln (3^{\frac{1}{4}})$ .

Tap for more steps...

Step 9.12.1

Multiply $- 1$ by $- 1$ .

$\frac{9}{3^{\frac{3}{4}}} + e^{1 \ln (3^{\frac{1}{4}})}$

Step 9.12.2

Multiply $\ln (3^{\frac{1}{4}})$ by $1$ .

$\frac{9}{3^{\frac{3}{4}}} + e^{\ln (3^{\frac{1}{4}})}$

Step 9.13

Exponentiation and log are inverse functions.

$\frac{9}{3^{\frac{3}{4}}} + 3^{\frac{1}{4}}$

Step 10

$x = - \frac{\ln (3)}{4}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - \frac{\ln (3)}{4}$ is a local minimum

Step 11

Find the y-value when $x = - \frac{\ln (3)}{4}$ .

Tap for more steps...

Step 11.1

Simplify $- \frac{\ln (3)}{4}$ to substitute in $f (x) = e^{3 x} + e^{- x}$ .

Tap for more steps...

Step 11.1.1

Rewrite $\frac{\ln (3)}{4}$ as $\frac{1}{4} \ln (3)$ .

$y = - (\frac{1}{4} \cdot \ln (3))$

Step 11.1.2

Simplify $\frac{1}{4} \ln (3)$ by moving $\frac{1}{4}$ inside the logarithm.

$y = - \ln (3^{\frac{1}{4}})$

Step 11.2

Replace the variable $x$ with $- \ln (3^{\frac{1}{4}})$ in the expression.

$f (- \ln (3^{\frac{1}{4}})) = e^{3 (- \ln (3^{\frac{1}{4}}))} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3

Simplify the result.

Tap for more steps...

Step 11.3.1

Simplify each term.

Tap for more steps...

Step 11.3.1.1

Multiply $3 (- \ln (3^{\frac{1}{4}}))$ .

Tap for more steps...

Step 11.3.1.1.1

Multiply $- 1$ by $3$ .

$f (- \ln (3^{\frac{1}{4}})) = e^{- 3 \ln (3^{\frac{1}{4}})} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.1.2

Simplify $- 3 \ln (3^{\frac{1}{4}})$ by moving $3$ inside the logarithm.

$f (- \ln (3^{\frac{1}{4}})) = e^{- \ln ({(3^{\frac{1}{4}})}^{3})} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.2

Simplify $- \ln ({(3^{\frac{1}{4}})}^{3})$ by moving $- 1$ inside the logarithm.

$f (- \ln (3^{\frac{1}{4}})) = e^{\ln ({({(3^{\frac{1}{4}})}^{3})}^{- 1})} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.3

Exponentiation and log are inverse functions.

$f (- \ln (3^{\frac{1}{4}})) = {({(3^{\frac{1}{4}})}^{3})}^{- 1} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.4

Multiply the exponents in ${({(3^{\frac{1}{4}})}^{3})}^{- 1}$ .

Tap for more steps...

Step 11.3.1.4.1

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

$f (- \ln (3^{\frac{1}{4}})) = {(3^{\frac{1}{4}})}^{3 \cdot - 1} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.4.2

Multiply $3$ by $- 1$ .

$f (- \ln (3^{\frac{1}{4}})) = {(3^{\frac{1}{4}})}^{- 3} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.5

Multiply the exponents in ${(3^{\frac{1}{4}})}^{- 3}$ .

Tap for more steps...

Step 11.3.1.5.1

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

$f (- \ln (3^{\frac{1}{4}})) = 3^{\frac{1}{4} \cdot - 3} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.5.2

Combine $\frac{1}{4}$ and $- 3$ .

$f (- \ln (3^{\frac{1}{4}})) = 3^{\frac{- 3}{4}} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.5.3

Move the negative in front of the fraction.

$f (- \ln (3^{\frac{1}{4}})) = 3^{- \frac{3}{4}} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.6

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

$f (- \ln (3^{\frac{1}{4}})) = \frac{1}{3^{\frac{3}{4}}} + e^{- (- \ln (3^{\frac{1}{4}}))}$

Step 11.3.1.7

Multiply $- (- \ln (3^{\frac{1}{4}}))$ .

Tap for more steps...

Step 11.3.1.7.1

Multiply $- 1$ by $- 1$ .

$f (- \ln (3^{\frac{1}{4}})) = \frac{1}{3^{\frac{3}{4}}} + e^{1 \ln (3^{\frac{1}{4}})}$

Step 11.3.1.7.2

Multiply $\ln (3^{\frac{1}{4}})$ by $1$ .

$f (- \ln (3^{\frac{1}{4}})) = \frac{1}{3^{\frac{3}{4}}} + e^{\ln (3^{\frac{1}{4}})}$

Step 11.3.1.8

Exponentiation and log are inverse functions.

$f (- \ln (3^{\frac{1}{4}})) = \frac{1}{3^{\frac{3}{4}}} + 3^{\frac{1}{4}}$

Step 11.3.2

The final answer is $\frac{1}{3^{\frac{3}{4}}} + 3^{\frac{1}{4}}$ .

$y = \frac{1}{3^{\frac{3}{4}}} + 3^{\frac{1}{4}}$

Step 12

These are the local extrema for $f (x) = e^{3 x} + e^{- x}$ .

$(- \frac{\ln (3)}{4}, \frac{1}{3^{\frac{3}{4}}} + 3^{\frac{1}{4}})$ is a local minima

Step 13