Calculus Examples

$f (x) = x - 5 \ln (x)$ , $\frac{1}{5} \leq x \leq 10$

Step 1

Find the critical points.

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Find the first derivative.

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Find the first derivative.

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Differentiate.

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

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

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

$f' (x) = 1 + \frac{d}{d x} (- 5 \ln (x))$

Evaluate $\frac{d}{d x} [- 5 \ln (x)]$ .

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

$f' (x) = 1 - 5 \frac{d}{d x} \ln (x)$

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$f' (x) = 1 - 5 \frac{1}{x}$

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

$f' (x) = 1 + \frac{- 5}{x}$

Move the negative in front of the fraction.

$f' (x) = 1 - \frac{5}{x}$

Reorder terms.

$f' (x) = - \frac{5}{x} + 1$

The first derivative of $f (x)$ with respect to $x$ is $- \frac{5}{x} + 1$ .

$- \frac{5}{x} + 1$

Set the first derivative equal to $0$ then solve the equation $- \frac{5}{x} + 1 = 0$ .

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Set the first derivative equal to $0$ .

$- \frac{5}{x} + 1 = 0$

Subtract $1$ from both sides of the equation.

$- \frac{5}{x} = - 1$

Find the LCD of the terms in the equation.

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

$x, 1$

The LCM of one and any expression is the expression.

$x$

Multiply each term in $- \frac{5}{x} = - 1$ by $x$ to eliminate the fractions.

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Multiply each term in $- \frac{5}{x} = - 1$ by $x$ .

$- \frac{5}{x} x = - x$

Simplify the left side.

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Cancel the common factor of $x$ .

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Move the leading negative in $- \frac{5}{x}$ into the numerator.

$\frac{- 5}{x} x = - x$

Cancel the common factor.

$\frac{- 5}{x} x = - x$

Rewrite the expression.

$- 5 = - x$

Solve the equation.

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Rewrite the equation as $- x = - 5$ .

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

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Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

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Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

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Divide $- 5$ by $- 1$ .

$x = 5$

Find the values where the derivative is undefined.

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Set the denominator in $\frac{5}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Evaluate $x - 5 \ln (x)$ at each $x$ value where the derivative is $0$ or undefined.

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Evaluate at $x = 5$ .

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Substitute $5$ for $x$ .

$(5) - 5 \ln (5)$

Simplify each term.

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Simplify $- 5 \ln (5)$ by moving $5$ inside the logarithm.

$5 - \ln (5^{5})$

Raise $5$ to the power of $5$ .

$5 - \ln (3125)$

Evaluate at $x = 0$ .

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Substitute $0$ for $x$ .

$(0) - 5 \ln (0)$

The natural logarithm of zero is undefined.

Undefined

List all of the points.

$(5, 5 - \ln (3125))$

Step 2

Evaluate at the included endpoints.

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Evaluate at $x = \frac{1}{5}$ .

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Substitute $\frac{1}{5}$ for $x$ .

$(\frac{1}{5}) - 5 \ln (\frac{1}{5})$

Simplify each term.

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

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

Apply the product rule to $\frac{1}{5}$ .

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

One to any power is one.

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

Raise $5$ to the power of $5$ .

$\frac{1}{5} - \ln (\frac{1}{3125})$

Evaluate at $x = 10$ .

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Substitute $10$ for $x$ .

$(10) - 5 \ln (10)$

Simplify each term.

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Simplify $- 5 \ln (10)$ by moving $5$ inside the logarithm.

$10 - \ln (10^{5})$

Raise $10$ to the power of $5$ .

$10 - \ln (100000)$

List all of the points.

$(\frac{1}{5}, \frac{1}{5} - \ln (\frac{1}{3125})), (10, 10 - \ln (100000))$

Step 3

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

Absolute Maximum: $(\frac{1}{5}, \frac{1}{5} - \ln (\frac{1}{3125}))$

Absolute Minimum: $(5, 5 - \ln (3125))$

Step 4