Calculus Examples

$f (x) = \ln (x) - x$ , $x > 0$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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

$\frac{d}{d x} [\ln (x)] + \frac{d}{d x} [- x]$

Step 1.1.1.2

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

$\frac{1}{x} + \frac{d}{d x} [- x]$

Step 1.1.1.3

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

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Step 1.1.1.3.1

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

$\frac{1}{x} - \frac{d}{d x} [x]$

Step 1.1.1.3.2

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

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

Step 1.1.1.3.3

Multiply $- 1$ by $1$ .

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

Set the first derivative equal to $0$ .

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

Step 1.2.2

Add $1$ to both sides of the equation.

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

Step 1.2.3

Find the LCD of the terms in the equation.

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Step 1.2.3.1

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

$x, 1$

Step 1.2.3.2

The LCM of one and any expression is the expression.

$x$

Step 1.2.4

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

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Step 1.2.4.1

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

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

Step 1.2.4.2

Simplify the left side.

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Step 1.2.4.2.1

Cancel the common factor of $x$ .

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Step 1.2.4.2.1.1

Cancel the common factor.

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

Step 1.2.4.2.1.2

Rewrite the expression.

$1 = 1 x$

Step 1.2.4.3

Simplify the right side.

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Step 1.2.4.3.1

Multiply $x$ by $1$ .

$1 = x$

Step 1.2.5

Rewrite the equation as $x = 1$ .

$x = 1$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

Set the denominator in $\frac{1}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 1.4

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

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Step 1.4.1

Evaluate at $x = 1$ .

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Step 1.4.1.1

Substitute $1$ for $x$ .

$\ln (1) - (1)$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

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

$0 - (1)$

Step 1.4.1.2.1.2

Multiply $- 1$ by $1$ .

$0 - 1$

Step 1.4.1.2.2

Subtract $1$ from $0$ .

$- 1$

Step 1.4.2

Evaluate at $x = 0$ .

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Step 1.4.2.1

Substitute $0$ for $x$ .

$\ln (0) - (0)$

Step 1.4.2.2

The natural logarithm of zero is undefined.

Undefined

Step 1.4.3

List all of the points.

$(1, - 1)$

Step 2

Use the first derivative test to determine which points can be maxima or minima.

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Step 2.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 1) \cup (1, \infty)$

Step 2.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 1)$ in the first derivative $\frac{1}{x} - 1$ to check if the result is negative or positive.

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Step 2.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = \frac{1}{- 2} - 1$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Move the negative in front of the fraction.

$f' (- 2) = - \frac{1}{2} - 1$

Step 2.2.2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (- 2) = - \frac{1}{2} - 1 \cdot \frac{2}{2}$

Step 2.2.2.3

Combine $- 1$ and $\frac{2}{2}$ .

$f' (- 2) = - \frac{1}{2} + \frac{- 1 \cdot 2}{2}$

Step 2.2.2.4

Combine the numerators over the common denominator.

$f' (- 2) = \frac{- 1 - 1 \cdot 2}{2}$

Step 2.2.2.5

Simplify the numerator.

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Step 2.2.2.5.1

Multiply $- 1$ by $2$ .

$f' (- 2) = \frac{- 1 - 2}{2}$

Step 2.2.2.5.2

Subtract $2$ from $- 1$ .

$f' (- 2) = \frac{- 3}{2}$

Step 2.2.2.6

Move the negative in front of the fraction.

$f' (- 2) = - \frac{3}{2}$

Step 2.2.2.7

The final answer is $- \frac{3}{2}$ .

$- \frac{3}{2}$

Step 2.3

Substitute any number, such as $4$ , from the interval $(1, \infty)$ in the first derivative $\frac{1}{x} - 1$ to check if the result is negative or positive.

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Step 2.3.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = \frac{1}{4} - 1$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 2.3.2.2

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

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

Step 2.3.2.3

Combine the numerators over the common denominator.

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

Step 2.3.2.4

Simplify the numerator.

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Step 2.3.2.4.1

Multiply $- 1$ by $4$ .

$f' (4) = \frac{1 - 4}{4}$

Step 2.3.2.4.2

Subtract $4$ from $1$ .

$f' (4) = \frac{- 3}{4}$

Step 2.3.2.5

Move the negative in front of the fraction.

$f' (4) = - \frac{3}{4}$

Step 2.3.2.6

The final answer is $- \frac{3}{4}$ .

$- \frac{3}{4}$

Step 2.4

Since the first derivative did not change signs around $x = 1$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 2.5

No local maxima or minima found for $f (x) = \ln (x) - x$ .

No local maxima or minima

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.

No absolute maximum

No absolute minimum

Step 4