Calculus Examples

$y = e^{x} - 3 e^{- x} - 4 x$

Step 1

Write $y = e^{x} - 3 e^{- x} - 4 x$ as a function.

$f (x) = e^{x} - 3 e^{- x} - 4 x$

Step 2

Find the first derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $e^{x} - 3 e^{- x} - 4 x$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 3 e^{- x}] + \frac{d}{d x} [- 4 x]$ .

$\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 3 e^{- x}] + \frac{d}{d x} [- 4 x]$

Step 2.2

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

$e^{x} + \frac{d}{d x} [- 3 e^{- x}] + \frac{d}{d x} [- 4 x]$

Step 2.3

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

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Step 2.3.1

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

$e^{x} - 3 \frac{d}{d x} [e^{- x}] + \frac{d}{d x} [- 4 x]$

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 $g (x) = - x$ .

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Step 2.3.2.1

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

$e^{x} - 3 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + \frac{d}{d x} [- 4 x]$

Step 2.3.2.2

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

$e^{x} - 3 (e^{u} \frac{d}{d x} [- x]) + \frac{d}{d x} [- 4 x]$

Step 2.3.2.3

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

$e^{x} - 3 (e^{- x} \frac{d}{d x} [- x]) + \frac{d}{d x} [- 4 x]$

Step 2.3.3

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

$e^{x} - 3 (e^{- x} (- \frac{d}{d x} [x])) + \frac{d}{d x} [- 4 x]$

Step 2.3.4

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

$e^{x} - 3 (e^{- x} (- 1 \cdot 1)) + \frac{d}{d x} [- 4 x]$

Step 2.3.5

Multiply $- 1$ by $1$ .

$e^{x} - 3 (e^{- x} \cdot - 1) + \frac{d}{d x} [- 4 x]$

Step 2.3.6

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

$e^{x} - 3 (- 1 \cdot e^{- x}) + \frac{d}{d x} [- 4 x]$

Step 2.3.7

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

$e^{x} - 3 (- e^{- x}) + \frac{d}{d x} [- 4 x]$

Step 2.3.8

Multiply $- 1$ by $- 3$ .

$e^{x} + 3 e^{- x} + \frac{d}{d x} [- 4 x]$

Step 2.4

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

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Step 2.4.1

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

$e^{x} + 3 e^{- x} - 4 \frac{d}{d x} [x]$

Step 2.4.2

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

$e^{x} + 3 e^{- x} - 4 \cdot 1$

Step 2.4.3

Multiply $- 4$ by $1$ .

$e^{x} + 3 e^{- x} - 4$

Step 2.5

Reorder terms.

$3 e^{- x} + e^{x} - 4$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $3 e^{- x} + e^{x} - 4$ with respect to $x$ is $\frac{d}{d x} [3 e^{- x}] + \frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 4]$ .

$f'' (x) = \frac{d}{d x} (3 e^{- x}) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.2

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

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Step 3.2.1

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

$f'' (x) = 3 \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

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 $g (x) = - x$ .

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Step 3.2.2.1

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

$f'' (x) = 3 (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x)) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.2.2.2

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

$f'' (x) = 3 (e^{u} \frac{d}{d x} (- x)) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.2.2.3

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

$f'' (x) = 3 (e^{- x} \frac{d}{d x} (- x)) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.2.3

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

$f'' (x) = 3 (e^{- x} (- \frac{d}{d x} x)) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.2.4

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

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

Step 3.2.5

Multiply $- 1$ by $1$ .

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

Step 3.2.6

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

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

Step 3.2.7

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

$f'' (x) = 3 (- e^{- x}) + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.2.8

Multiply $- 1$ by $3$ .

$f'' (x) = - 3 e^{- x} + \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 4)$

Step 3.3

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

$f'' (x) = - 3 e^{- x} + e^{x} + \frac{d}{d x} (- 4)$

Step 3.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$f'' (x) = - 3 e^{- x} + e^{x} + 0$

Step 3.5

Simplify.

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Step 3.5.1

Add $- 3 e^{- x} + e^{x}$ and $0$ .

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

Step 3.5.2

Reorder terms.

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

Step 4

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

$3 e^{- x} + e^{x} - 4 = 0$

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 $e^{x} - 3 e^{- x} - 4 x$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 3 e^{- x}] + \frac{d}{d x} [- 4 x]$ .

$f' (x) = \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 3 e^{- x}) + \frac{d}{d x} (- 4 x)$

Step 5.1.2

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

$f' (x) = e^{x} + \frac{d}{d x} (- 3 e^{- x}) + \frac{d}{d x} (- 4 x)$

Step 5.1.3

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

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Step 5.1.3.1

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

$f' (x) = e^{x} - 3 \frac{d}{d x} e^{- x} + \frac{d}{d x} (- 4 x)$

Step 5.1.3.2

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 5.1.3.2.1

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

$f' (x) = e^{x} - 3 (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x)) + \frac{d}{d x} (- 4 x)$

Step 5.1.3.2.2

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

$f' (x) = e^{x} - 3 (e^{u} \frac{d}{d x} (- x)) + \frac{d}{d x} (- 4 x)$

Step 5.1.3.2.3

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

$f' (x) = e^{x} - 3 (e^{- x} \frac{d}{d x} (- x)) + \frac{d}{d x} (- 4 x)$

Step 5.1.3.3

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

$f' (x) = e^{x} - 3 (e^{- x} (- \frac{d}{d x} x)) + \frac{d}{d x} (- 4 x)$

Step 5.1.3.4

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

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

Step 5.1.3.5

Multiply $- 1$ by $1$ .

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

Step 5.1.3.6

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

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

Step 5.1.3.7

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

$f' (x) = e^{x} - 3 (- e^{- x}) + \frac{d}{d x} (- 4 x)$

Step 5.1.3.8

Multiply $- 1$ by $- 3$ .

$f' (x) = e^{x} + 3 e^{- x} + \frac{d}{d x} (- 4 x)$

Step 5.1.4

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

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Step 5.1.4.1

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

$f' (x) = e^{x} + 3 e^{- x} - 4 \frac{d}{d x} x$

Step 5.1.4.2

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

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

Step 5.1.4.3

Multiply $- 4$ by $1$ .

$f' (x) = e^{x} + 3 e^{- x} - 4$

Step 5.1.5

Reorder terms.

$f' (x) = 3 e^{- x} + e^{x} - 4$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $3 e^{- x} + e^{x} - 4$ .

$3 e^{- x} + e^{x} - 4$

Step 6

Set the first derivative equal to $0$ then solve the equation $3 e^{- x} + e^{x} - 4 = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$3 e^{- x} + e^{x} - 4 = 0$

Step 6.2

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

$3 {(e^{x})}^{- 1} + e^{x} - 4 = 0$

Step 6.3

Substitute $u$ for $e^{x}$ .

$3 u^{- 1} + u - 4 = 0$

Step 6.4

Simplify each term.

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Step 6.4.1

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

$3 (\frac{1}{u}) + u - 4 = 0$

Step 6.4.2

Combine $3$ and $\frac{1}{u}$ .

$\frac{3}{u} + u - 4 = 0$

Step 6.5

Reorder $\frac{3}{u}$ and $u$ .

$u + \frac{3}{u} - 4 = 0$

Step 6.6

Solve for $u$ .

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Step 6.6.1

Find the LCD of the terms in the equation.

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Step 6.6.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, u, 1, 1$

Step 6.6.1.2

The LCM of one and any expression is the expression.

$u$

Step 6.6.2

Multiply each term in $u + \frac{3}{u} - 4 = 0$ by $u$ to eliminate the fractions.

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Step 6.6.2.1

Multiply each term in $u + \frac{3}{u} - 4 = 0$ by $u$ .

$u \cdot u + \frac{3}{u} u - 4 u = 0 u$

Step 6.6.2.2

Simplify the left side.

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Step 6.6.2.2.1

Simplify each term.

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Step 6.6.2.2.1.1

Multiply $u$ by $u$ .

$u^{2} + \frac{3}{u} u - 4 u = 0 u$

Step 6.6.2.2.1.2

Cancel the common factor of $u$ .

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Step 6.6.2.2.1.2.1

Cancel the common factor.

$u^{2} + \frac{3}{u} u - 4 u = 0 u$

Step 6.6.2.2.1.2.2

Rewrite the expression.

$u^{2} + 3 - 4 u = 0 u$

Step 6.6.2.3

Simplify the right side.

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Step 6.6.2.3.1

Multiply $0$ by $u$ .

$u^{2} + 3 - 4 u = 0$

Step 6.6.3

Solve the equation.

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Step 6.6.3.1

Factor $u^{2} + 3 - 4 u$ using the AC method.

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Step 6.6.3.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Step 6.6.3.1.2

Write the factored form using these integers.

$(u - 3) (u - 1) = 0$

Step 6.6.3.2

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

$u - 3 = 0$

$u - 1 = 0$

Step 6.6.3.3

Set $u - 3$ equal to $0$ and solve for $u$ .

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Step 6.6.3.3.1

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Step 6.6.3.3.2

Add $3$ to both sides of the equation.

$u = 3$

Step 6.6.3.4

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 6.6.3.4.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 6.6.3.4.2

Add $1$ to both sides of the equation.

$u = 1$

Step 6.6.3.5

The final solution is all the values that make $(u - 3) (u - 1) = 0$ true.

$u = 3, 1$

Step 6.7

Substitute $3$ for $u$ in $u = e^{x}$ .

$3 = e^{x}$

Step 6.8

Solve $3 = e^{x}$ .

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Step 6.8.1

Rewrite the equation as $e^{x} = 3$ .

$e^{x} = 3$

Step 6.8.2

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

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

Step 6.8.3

Expand the left side.

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Step 6.8.3.1

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

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

Step 6.8.3.2

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

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

Step 6.8.3.3

Multiply $x$ by $1$ .

$x = \ln (3)$

Step 6.9

Substitute $1$ for $u$ in $u = e^{x}$ .

$1 = e^{x}$

Step 6.10

Solve $1 = e^{x}$ .

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Step 6.10.1

Rewrite the equation as $e^{x} = 1$ .

$e^{x} = 1$

Step 6.10.2

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

$\ln (e^{x}) = \ln (1)$

Step 6.10.3

Expand the left side.

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Step 6.10.3.1

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

$x \ln (e) = \ln (1)$

Step 6.10.3.2

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

$x \cdot 1 = \ln (1)$

Step 6.10.3.3

Multiply $x$ by $1$ .

$x = \ln (1)$

Step 6.10.4

The natural logarithm of $1$ is $0$ .

$x = 0$

Step 6.11

List the solutions that makes the equation true.

$x = \ln (3), 0$

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.

$x = \ln (3), 0$

Step 9

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

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

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

Exponentiation and log are inverse functions.

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

Step 10.1.2

Simplify $- (\ln (3))$ by moving $- 1$ inside the logarithm.

$3 - 3 e^{\ln (3^{- 1})}$

Step 10.1.3

Exponentiation and log are inverse functions.

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

Step 10.1.4

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

$3 - 3 \cdot \frac{1}{3}$

Step 10.1.5

Cancel the common factor of $3$ .

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Step 10.1.5.1

Factor $3$ out of $- 3$ .

$3 + 3 (- 1) \cdot \frac{1}{3}$

Step 10.1.5.2

Cancel the common factor.

$3 + 3 \cdot - 1 \cdot \frac{1}{3}$

Step 10.1.5.3

Rewrite the expression.

$3 - 1$

Step 10.2

Subtract $1$ from $3$ .

$2$

Step 11

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

$x = \ln (3)$ is a local minimum

Step 12

Find the y-value when $x = \ln (3)$ .

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Step 12.1

Replace the variable $x$ with $\ln (3)$ in the expression.

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

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Exponentiation and log are inverse functions.

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

Step 12.2.1.2

Simplify $- (\ln (3))$ by moving $- 1$ inside the logarithm.

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

Step 12.2.1.3

Exponentiation and log are inverse functions.

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

Step 12.2.1.4

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

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

Step 12.2.1.5

Cancel the common factor of $3$ .

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Step 12.2.1.5.1

Factor $3$ out of $- 3$ .

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

Step 12.2.1.5.2

Cancel the common factor.

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

Step 12.2.1.5.3

Rewrite the expression.

$f (\ln (3)) = 3 - 1 - 4 \ln (3)$

Step 12.2.1.6

Simplify $- 4 \ln (3)$ by moving $4$ inside the logarithm.

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

Step 12.2.1.7

Raise $3$ to the power of $4$ .

$f (\ln (3)) = 3 - 1 - \ln (81)$

Step 12.2.2

Subtract $1$ from $3$ .

$f (\ln (3)) = 2 - \ln (81)$

Step 12.2.3

The final answer is $2 - \ln (81)$ .

$y = 2 - \ln (81)$

Step 13

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$e^{0} - 3 e^{- (0)}$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

Anything raised to $0$ is $1$ .

$1 - 3 e^{- (0)}$

Step 14.1.2

Multiply $- 1$ by $0$ .

$1 - 3 e^{0}$

Step 14.1.3

Anything raised to $0$ is $1$ .

$1 - 3 \cdot 1$

Step 14.1.4

Multiply $- 3$ by $1$ .

$1 - 3$

Step 14.2

Subtract $3$ from $1$ .

$- 2$

Step 15

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 16

Find the y-value when $x = 0$ .

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Step 16.1

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

$f (0) = e^{0} - 3 e^{- (0)} - 4 \cdot 0$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

Anything raised to $0$ is $1$ .

$f (0) = 1 - 3 e^{- (0)} - 4 \cdot 0$

Step 16.2.1.2

Multiply $- 1$ by $0$ .

$f (0) = 1 - 3 e^{0} - 4 \cdot 0$

Step 16.2.1.3

Anything raised to $0$ is $1$ .

$f (0) = 1 - 3 \cdot 1 - 4 \cdot 0$

Step 16.2.1.4

Multiply $- 3$ by $1$ .

$f (0) = 1 - 3 - 4 \cdot 0$

Step 16.2.1.5

Multiply $- 4$ by $0$ .

$f (0) = 1 - 3 + 0$

Step 16.2.2

Simplify by adding and subtracting.

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Step 16.2.2.1

Subtract $3$ from $1$ .

$f (0) = - 2 + 0$

Step 16.2.2.2

Add $- 2$ and $0$ .

$f (0) = - 2$

Step 16.2.3

The final answer is $- 2$ .

$y = - 2$

Step 17

These are the local extrema for $f (x) = e^{x} - 3 e^{- x} - 4 x$ .

$(\ln (3), 2 - \ln (81))$ is a local minima

$(0, - 2)$ is a local maxima

Step 18