Algebra Examples

$f (x) = e^{0.5 x} + 64 e^{- 0.5 x}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $e^{0.5 x} + 64 e^{- 0.5 x}$ with respect to $x$ is $\frac{d}{d x} [e^{0.5 x}] + \frac{d}{d x} [64 e^{- 0.5 x}]$ .

$f' (x) = \frac{d}{d x} (e^{0.5 x}) + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2

Evaluate $\frac{d}{d x} [e^{0.5 x}]$ .

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Step 1.1.2.1

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 $g (x) = 0.5 x$ .

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Step 1.1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $0.5 x$ .

$f' (x) = \frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (0.5 x) + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2.1.2

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

$f' (x) = e^{u_{1}} \frac{d}{d x} (0.5 x) + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2.1.3

Replace all occurrences of $u_{1}$ with $0.5 x$ .

$f' (x) = e^{0.5 x} \frac{d}{d x} (0.5 x) + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2.2

Since $0.5$ is constant with respect to $x$ , the derivative of $0.5 x$ with respect to $x$ is $0.5 \frac{d}{d x} [x]$ .

$f' (x) = e^{0.5 x} (0.5 \frac{d}{d x} (x)) + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2.3

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

$f' (x) = e^{0.5 x} (0.5 \cdot 1) + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2.4

Multiply $0.5$ by $1$ .

$f' (x) = e^{0.5 x} \cdot 0.5 + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.2.5

Move $0.5$ to the left of $e^{0.5 x}$ .

$f' (x) = 0.5 e^{0.5 x} + \frac{d}{d x} (64 e^{- 0.5 x})$

Step 1.1.3

Evaluate $\frac{d}{d x} [64 e^{- 0.5 x}]$ .

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Step 1.1.3.1

Since $64$ is constant with respect to $x$ , the derivative of $64 e^{- 0.5 x}$ with respect to $x$ is $64 \frac{d}{d x} [e^{- 0.5 x}]$ .

$f' (x) = 0.5 e^{0.5 x} + 64 \frac{d}{d x} (e^{- 0.5 x})$

Step 1.1.3.2

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 $g (x) = - 0.5 x$ .

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Step 1.1.3.2.1

To apply the Chain Rule, set $u_{2}$ as $- 0.5 x$ .

$f' (x) = 0.5 e^{0.5 x} + 64 (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- 0.5 x))$

Step 1.1.3.2.2

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

$f' (x) = 0.5 e^{0.5 x} + 64 (e^{u_{2}} \frac{d}{d x} (- 0.5 x))$

Step 1.1.3.2.3

Replace all occurrences of $u_{2}$ with $- 0.5 x$ .

$f' (x) = 0.5 e^{0.5 x} + 64 (e^{- 0.5 x} \frac{d}{d x} (- 0.5 x))$

Step 1.1.3.3

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x]$ .

$f' (x) = 0.5 e^{0.5 x} + 64 (e^{- 0.5 x} (- 0.5 \frac{d}{d x} x))$

Step 1.1.3.4

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

$f' (x) = 0.5 e^{0.5 x} + 64 (e^{- 0.5 x} (- 0.5 \cdot 1))$

Step 1.1.3.5

Multiply $- 0.5$ by $1$ .

$f' (x) = 0.5 e^{0.5 x} + 64 (e^{- 0.5 x} \cdot - 0.5)$

Step 1.1.3.6

Move $- 0.5$ to the left of $e^{- 0.5 x}$ .

$f' (x) = 0.5 e^{0.5 x} + 64 (- 0.5 \cdot e^{- 0.5 x})$

Step 1.1.3.7

Multiply $- 0.5$ by $64$ .

$f' (x) = 0.5 e^{0.5 x} - 32 e^{- 0.5 x}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $0.5 e^{0.5 x} - 32 e^{- 0.5 x}$ .

$0.5 e^{0.5 x} - 32 e^{- 0.5 x}$

Step 2

Set the first derivative equal to $0$ then solve the equation $0.5 e^{0.5 x} - 32 e^{- 0.5 x} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$0.5 e^{0.5 x} - 32 e^{- 0.5 x} = 0$

Step 2.2

Move $- 32 e^{- 0.5 x}$ to the right side of the equation by adding it to both sides.

$0.5 e^{0.5 x} = 32 e^{- 0.5 x}$

Step 2.3

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

$\ln (0.5 e^{0.5 x}) = \ln (32 e^{- 0.5 x})$

Step 2.4

Expand the left side.

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Step 2.4.1

Rewrite $\ln (0.5 e^{0.5 x})$ as $\ln (0.5) + \ln (e^{0.5 x})$ .

$\ln (0.5) + \ln (e^{0.5 x}) = \ln (32 e^{- 0.5 x})$

Step 2.4.2

Expand $\ln (e^{0.5 x})$ by moving $0.5 x$ outside the logarithm.

$\ln (0.5) + 0.5 x \ln (e) = \ln (32 e^{- 0.5 x})$

Step 2.4.3

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

$\ln (0.5) + 0.5 x \cdot 1 = \ln (32 e^{- 0.5 x})$

Step 2.4.4

Multiply $0.5$ by $1$ .

$\ln (0.5) + 0.5 x = \ln (32 e^{- 0.5 x})$

Step 2.5

Expand the right side.

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Step 2.5.1

Rewrite $\ln (32 e^{- 0.5 x})$ as $\ln (32) + \ln (e^{- 0.5 x})$ .

$\ln (0.5) + 0.5 x = \ln (32) + \ln (e^{- 0.5 x})$

Step 2.5.2

Expand $\ln (e^{- 0.5 x})$ by moving $- 0.5 x$ outside the logarithm.

$\ln (0.5) + 0.5 x = \ln (32) - 0.5 x \ln (e)$

Step 2.5.3

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

$\ln (0.5) + 0.5 x = \ln (32) - 0.5 x \cdot 1$

Step 2.5.4

Multiply $- 0.5$ by $1$ .

$\ln (0.5) + 0.5 x = \ln (32) - 0.5 x$

Step 2.6

Move all the terms containing a logarithm to the left side of the equation.

$\ln (0.5) - \ln (32) = - 0.5 x - 0.5 x$

Step 2.7

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{0.5}{32}) = - 0.5 x - 0.5 x$

Step 2.8

Divide $0.5$ by $32$ .

$\ln (0.015625) = - 0.5 x - 0.5 x$

Step 2.9

Subtract $0.5 x$ from $- 0.5 x$ .

$\ln (0.015625) = - 1 x$

Step 2.10

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- 1 x = \ln (0.015625)$

Step 2.11

Divide each term in $- 1 x = \ln (0.015625)$ by $- 1$ and simplify.

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Step 2.11.1

Divide each term in $- 1 x = \ln (0.015625)$ by $- 1$ .

$\frac{- 1 x}{- 1} = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2

Simplify the left side.

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Step 2.11.2.1

Cancel the common factor of $- 1$ .

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Step 2.11.2.1.1

Rewrite $- 1$ as $- 1 (1)$ .

$\frac{- 1 (1) x}{- 1} = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.1.2

Factor $1$ out of $- 1 (1) x$ .

$\frac{1 (- 1 \cdot 1 x)}{- 1} = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.1.3

Move the negative one from the denominator of $\frac{- 1 \cdot 1 x}{- 1}$ .

$- 1 \cdot (- 1 \cdot 1 x) = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.2

Simplify the expression.

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Step 2.11.2.2.1

Rewrite $- 1 \cdot (- 1 \cdot 1 x)$ as $- (- 1 \cdot 1 x)$ .

$- (- 1 \cdot 1 x) = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.2.2

Multiply $- 1$ by $1$ .

$- (- 1 x) = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.2.3

Rewrite $- 1 x$ as $- x$ .

$- - x = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.3

Multiply $- - x$ .

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Step 2.11.2.3.1

Multiply $- 1$ by $- 1$ .

$1 x = \frac{\ln (0.015625)}{- 1}$

Step 2.11.2.3.2

Multiply $x$ by $1$ .

$x = \frac{\ln (0.015625)}{- 1}$

Step 2.11.3

Simplify the right side.

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Step 2.11.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (0.015625)}{1}$

Step 2.11.3.2

Divide $\ln (0.015625)$ by $1$ .

$x = - \ln (0.015625)$

Step 3

The values which make the derivative equal to $0$ are $- \ln (0.015625)$ .

$- \ln (0.015625)$

Step 4

After finding the point that makes the derivative $f' (x) = 0.5 e^{0.5 x} - 32 e^{- 0.5 x}$ equal to $0$ or undefined, the interval to check where $f (x) = e^{0.5 x} + 64 e^{- 0.5 x}$ is increasing and where it is decreasing is $(- \infty, - \ln (0.015625)) \cup (- \ln (0.015625), \infty)$ .

$(- \infty, - \ln (0.015625)) \cup (- \ln (0.015625), \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \ln (0.015625))$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $3.15888308$ in the expression.

$f' (3.15888308) = 0.5 e^{0.5 (3.15888308)} - 32 e^{- 0.5 \cdot 3.15888308}$

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

Multiply $0.5$ by $3.15888308$ .

$f' (3.15888308) = 0.5 e^{1.57944154} - 32 e^{- 0.5 \cdot 3.15888308}$

Step 5.2.1.2

Multiply $0.5$ by $e^{1.57944154}$ .

$f' (3.15888308) = 2.42612263 - 32 e^{- 0.5 \cdot 3.15888308}$

Step 5.2.1.3

Multiply $- 0.5$ by $3.15888308$ .

$f' (3.15888308) = 2.42612263 - 32 e^{- 1.57944154}$

Step 5.2.1.4

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

$f' (3.15888308) = 2.42612263 - 32 \frac{1}{e^{1.57944154}}$

Step 5.2.1.5

Combine $- 32$ and $\frac{1}{e^{1.57944154}}$ .

$f' (3.15888308) = 2.42612263 + \frac{- 32}{e^{1.57944154}}$

Step 5.2.1.6

Move the negative in front of the fraction.

$f' (3.15888308) = 2.42612263 - \frac{32}{e^{1.57944154}}$

Step 5.2.1.7

Replace $e$ with an approximation.

$f' (3.15888308) = 2.42612263 - \frac{32}{{2.71828182}^{1.57944154}}$

Step 5.2.1.8

Raise $2.71828182$ to the power of $1.57944154$ .

$f' (3.15888308) = 2.42612263 - \frac{32}{4.85224527}$

Step 5.2.1.9

Divide $32$ by $4.85224527$ .

$f' (3.15888308) = 2.42612263 - 1 \cdot 6.59488508$

Step 5.2.1.10

Multiply $- 1$ by $6.59488508$ .

$f' (3.15888308) = 2.42612263 - 6.59488508$

Step 5.2.2

Subtract $6.59488508$ from $2.42612263$ .

$f' (3.15888308) = - 4.16876244$

Step 5.2.3

The final answer is $- 4.16876244$ .

$- 4.16876244$

Step 5.3

At $x = 3.15888308$ the derivative is $- 4.16876244$ . Since this is negative, the function is decreasing on $(- \infty, - \ln (0.015625))$ .

Decreasing on $(- \infty, - \ln (0.015625))$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- \ln (0.015625), \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $5.15888308$ in the expression.

$f' (5.15888308) = 0.5 e^{0.5 (5.15888308)} - 32 e^{- 0.5 \cdot 5.15888308}$

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

Multiply $0.5$ by $5.15888308$ .

$f' (5.15888308) = 0.5 e^{2.57944154} - 32 e^{- 0.5 \cdot 5.15888308}$

Step 6.2.1.2

Multiply $0.5$ by $e^{2.57944154}$ .

$f' (5.15888308) = 6.59488508 - 32 e^{- 0.5 \cdot 5.15888308}$

Step 6.2.1.3

Multiply $- 0.5$ by $5.15888308$ .

$f' (5.15888308) = 6.59488508 - 32 e^{- 2.57944154}$

Step 6.2.1.4

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

$f' (5.15888308) = 6.59488508 - 32 \frac{1}{e^{2.57944154}}$

Step 6.2.1.5

Combine $- 32$ and $\frac{1}{e^{2.57944154}}$ .

$f' (5.15888308) = 6.59488508 + \frac{- 32}{e^{2.57944154}}$

Step 6.2.1.6

Move the negative in front of the fraction.

$f' (5.15888308) = 6.59488508 - \frac{32}{e^{2.57944154}}$

Step 6.2.1.7

Replace $e$ with an approximation.

$f' (5.15888308) = 6.59488508 - \frac{32}{{2.71828182}^{2.57944154}}$

Step 6.2.1.8

Raise $2.71828182$ to the power of $2.57944154$ .

$f' (5.15888308) = 6.59488508 - \frac{32}{13.18977016}$

Step 6.2.1.9

Divide $32$ by $13.18977016$ .

$f' (5.15888308) = 6.59488508 - 1 \cdot 2.42612263$

Step 6.2.1.10

Multiply $- 1$ by $2.42612263$ .

$f' (5.15888308) = 6.59488508 - 2.42612263$

Step 6.2.2

Subtract $2.42612263$ from $6.59488508$ .

$f' (5.15888308) = 4.16876244$

Step 6.2.3

The final answer is $4.16876244$ .

$4.16876244$

Step 6.3

At $x = 5.15888308$ the derivative is $4.16876244$ . Since this is positive, the function is increasing on $(- \ln (0.015625), \infty)$ .

Increasing on $(- \ln (0.015625), \infty)$ since $f' (x) > 0$

Step 7

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \ln (0.015625), \infty)$

Decreasing on: $(- \infty, - \ln (0.015625))$

Step 8