Algebra Examples

$e^{x} + e^{- 5 x}$

Step 1

Write $e^{x} + e^{- 5 x}$ as a function.

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

Step 2

Find the first derivative of the function.

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By the Sum Rule, the derivative of $e^{x} + e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [e^{- 5 x}]$ .

$\frac{d}{d x} [e^{x}] + \frac{d}{d x} [e^{- 5 x}]$

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

$e^{x} + \frac{d}{d x} [e^{- 5 x}]$

Evaluate $\frac{d}{d x} [e^{- 5 x}]$ .

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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 $g (x) = - 5 x$ .

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To apply the Chain Rule, set $u$ as $- 5 x$ .

$e^{x} + \frac{d}{d u} [e^{u}] \frac{d}{d x} [- 5 x]$

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

$e^{x} + e^{u} \frac{d}{d x} [- 5 x]$

Replace all occurrences of $u$ with $- 5 x$ .

$e^{x} + e^{- 5 x} \frac{d}{d x} [- 5 x]$

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

$e^{x} + e^{- 5 x} (- 5 \frac{d}{d x} [x])$

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

$e^{x} + e^{- 5 x} (- 5 \cdot 1)$

Multiply $- 5$ by $1$ .

$e^{x} + e^{- 5 x} \cdot - 5$

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

$e^{x} - 5 e^{- 5 x}$

Step 3

Find the second derivative of the function.

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By the Sum Rule, the derivative of $e^{x} - 5 e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 5 e^{- 5 x}]$ .

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

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

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

Evaluate $\frac{d}{d x} [- 5 e^{- 5 x}]$ .

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

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

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 $g (x) = - 5 x$ .

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To apply the Chain Rule, set $u$ as $- 5 x$ .

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

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

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

Replace all occurrences of $u$ with $- 5 x$ .

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

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

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

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

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

Multiply $- 5$ by $1$ .

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

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

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

Multiply $- 5$ by $- 5$ .

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

Reorder terms.

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

Step 4

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

$e^{x} - 5 e^{- 5 x} = 0$

Step 5

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $e^{x} + e^{- 5 x}$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [e^{- 5 x}]$ .

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

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

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

Evaluate $\frac{d}{d x} [e^{- 5 x}]$ .

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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 $g (x) = - 5 x$ .

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To apply the Chain Rule, set $u$ as $- 5 x$ .

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

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

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

Replace all occurrences of $u$ with $- 5 x$ .

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

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

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

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

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

Multiply $- 5$ by $1$ .

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

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

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

The first derivative of $f (x)$ with respect to $x$ is $e^{x} - 5 e^{- 5 x}$ .

$e^{x} - 5 e^{- 5 x}$

Step 6

Set the first derivative equal to $0$ then solve the equation $e^{x} - 5 e^{- 5 x} = 0$ .

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Set the first derivative equal to $0$ .

$e^{x} - 5 e^{- 5 x} = 0$

Move $- 5 e^{- 5 x}$ to the right side of the equation by adding it to both sides.

$e^{x} = 5 e^{- 5 x}$

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

$\ln (e^{x}) = \ln (5 e^{- 5 x})$

Expand the left side.

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

$x \ln (e) = \ln (5 e^{- 5 x})$

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

$x \cdot 1 = \ln (5 e^{- 5 x})$

Multiply $x$ by $1$ .

$x = \ln (5 e^{- 5 x})$

Expand the right side.

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Rewrite $\ln (5 e^{- 5 x})$ as $\ln (5) + \ln (e^{- 5 x})$ .

$x = \ln (5) + \ln (e^{- 5 x})$

Expand $\ln (e^{- 5 x})$ by moving $- 5 x$ outside the logarithm.

$x = \ln (5) - 5 x \ln (e)$

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

$x = \ln (5) - 5 x \cdot 1$

Multiply $- 5$ by $1$ .

$x = \ln (5) - 5 x$

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

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Add $5 x$ to both sides of the equation.

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

Add $x$ and $5 x$ .

$6 x = \ln (5)$

Divide each term in $6 x = \ln (5)$ by $6$ and simplify.

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Divide each term in $6 x = \ln (5)$ by $6$ .

$\frac{6 x}{6} = \frac{\ln (5)}{6}$

Simplify the left side.

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Cancel the common factor of $6$ .

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Cancel the common factor.

$\frac{6 x}{6} = \frac{\ln (5)}{6}$

Divide $x$ by $1$ .

$x = \frac{\ln (5)}{6}$

Step 7

Find the values where the derivative is undefined.

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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.

$x = \frac{\ln (5)}{6}$

Step 9

Evaluate the second derivative at $x = \frac{\ln (5)}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$25 e^{- 5 \frac{\ln (5)}{6}} + e^{\frac{\ln (5)}{6}}$

Step 10

Simplify each term.

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Rewrite $\frac{\ln (5)}{6}$ as $\frac{1}{6} \ln (5)$ .

$25 e^{- 5 (\frac{1}{6} \ln (5))} + e^{\frac{\ln (5)}{6}}$

Simplify $\frac{1}{6} \ln (5)$ by moving $\frac{1}{6}$ inside the logarithm.

$25 e^{- 5 \ln (5^{\frac{1}{6}})} + e^{\frac{\ln (5)}{6}}$

Simplify $- 5 \ln (5^{\frac{1}{6}})$ by moving $- 5$ inside the logarithm.

$25 e^{\ln ({(5^{\frac{1}{6}})}^{- 5})} + e^{\frac{\ln (5)}{6}}$

Exponentiation and log are inverse functions.

$25 {(5^{\frac{1}{6}})}^{- 5} + e^{\frac{\ln (5)}{6}}$

Multiply the exponents in ${(5^{\frac{1}{6}})}^{- 5}$ .

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

$25 \cdot 5^{\frac{1}{6} \cdot - 5} + e^{\frac{\ln (5)}{6}}$

Combine $\frac{1}{6}$ and $- 5$ .

$25 \cdot 5^{\frac{- 5}{6}} + e^{\frac{\ln (5)}{6}}$

Move the negative in front of the fraction.

$25 \cdot 5^{- \frac{5}{6}} + e^{\frac{\ln (5)}{6}}$

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

$25 \cdot \frac{1}{5^{\frac{5}{6}}} + e^{\frac{\ln (5)}{6}}$

Combine $25$ and $\frac{1}{5^{\frac{5}{6}}}$ .

$\frac{25}{5^{\frac{5}{6}}} + e^{\frac{\ln (5)}{6}}$

Rewrite $\frac{\ln (5)}{6}$ as $\frac{1}{6} \ln (5)$ .

$\frac{25}{5^{\frac{5}{6}}} + e^{\frac{1}{6} \ln (5)}$

Simplify $\frac{1}{6} \ln (5)$ by moving $\frac{1}{6}$ inside the logarithm.

$\frac{25}{5^{\frac{5}{6}}} + e^{\ln (5^{\frac{1}{6}})}$

Exponentiation and log are inverse functions.

$\frac{25}{5^{\frac{5}{6}}} + 5^{\frac{1}{6}}$

Step 11

$x = \frac{\ln (5)}{6}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{\ln (5)}{6}$ is a local minimum

Step 12

Find the y-value when $x = \frac{\ln (5)}{6}$ .

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Simplify $\frac{\ln (5)}{6}$ to substitute in $f (x) = e^{x} + e^{- 5 x}$ .

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Rewrite $\frac{\ln (5)}{6}$ as $\frac{1}{6} \ln (5)$ .

$y = \frac{1}{6} \cdot \ln (5)$

Simplify $\frac{1}{6} \ln (5)$ by moving $\frac{1}{6}$ inside the logarithm.

$y = \ln (5^{\frac{1}{6}})$

Replace the variable $x$ with $\ln (5^{\frac{1}{6}})$ in the expression.

$f (\ln (5^{\frac{1}{6}})) = e^{\ln (5^{\frac{1}{6}})} + e^{- 5 \ln (5^{\frac{1}{6}})}$

Simplify the result.

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Simplify each term.

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Exponentiation and log are inverse functions.

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + e^{- 5 \ln (5^{\frac{1}{6}})}$

Simplify $- 5 \ln (5^{\frac{1}{6}})$ by moving $- 5$ inside the logarithm.

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + e^{\ln ({(5^{\frac{1}{6}})}^{- 5})}$

Exponentiation and log are inverse functions.

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + {(5^{\frac{1}{6}})}^{- 5}$

Multiply the exponents in ${(5^{\frac{1}{6}})}^{- 5}$ .

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

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + 5^{\frac{1}{6} \cdot - 5}$

Combine $\frac{1}{6}$ and $- 5$ .

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + 5^{\frac{- 5}{6}}$

Move the negative in front of the fraction.

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + 5^{- \frac{5}{6}}$

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

$f (\ln (5^{\frac{1}{6}})) = 5^{\frac{1}{6}} + \frac{1}{5^{\frac{5}{6}}}$

The final answer is $5^{\frac{1}{6}} + \frac{1}{5^{\frac{5}{6}}}$ .

$y = 5^{\frac{1}{6}} + \frac{1}{5^{\frac{5}{6}}}$

Step 13

These are the local extrema for $f (x) = e^{x} + e^{- 5 x}$ .

$(\frac{\ln (5)}{6}, 5^{\frac{1}{6}} + \frac{1}{5^{\frac{5}{6}}})$ is a local minima

Step 14