Calculus Examples

$f (x) = \ln (x)$ , $[1, 8]$

Step 1

If $f$ is continuous on the interval $[a, b]$ and differentiable on $(a, b)$ , then at least one real number $c$ exists in the interval $(a, b)$ such that $f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at $x = c$ and the slope of the line through the points $(a, f (a))$ and $(b, f (b))$ .

If $f (x)$ is continuous on $[a, b]$

and if $f (x)$ differentiable on $(a, b)$ ,

then there exists at least one point, $c$ in $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

Check if $f (x) = \ln (x)$ is continuous.

Tap for more steps...

Step 2.1

To find whether the function is continuous on $[1, 8]$ or not, find the domain of $f (x) = \ln (x)$ .

Tap for more steps...

Step 2.1.1

Set the argument in $\ln (x)$ greater than $0$ to find where the expression is defined.

$x > 0$

Step 2.1.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 2.2

$f (x)$ is continuous on $[1, 8]$ .

The function is continuous.

Step 3

Find the derivative.

Tap for more steps...

Step 3.1

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

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

Step 3.2

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

$\frac{1}{x}$

Step 4

Find if the derivative is continuous on $(1, 8)$ .

Tap for more steps...

Step 4.1

To find whether the function is continuous on $(1, 8)$ or not, find the domain of $f' (x) = \frac{1}{x}$ .

Tap for more steps...

Step 4.1.1

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

$x = 0$

Step 4.1.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Step 4.2

$f' (x)$ is continuous on $(1, 8)$ .

The function is continuous.

Step 5

The function is differentiable on $(1, 8)$ because the derivative is continuous on $(1, 8)$ .

The function is differentiable.

Step 6

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[1, 8]$ and differentiable on $(1, 8)$ .

$f (x)$ is continuous on $[1, 8]$ and differentiable on $(1, 8)$ .

Step 7

Evaluate $f (a)$ from the interval $[1, 8]$ .

Tap for more steps...

Step 7.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln (1)$

Step 7.2

Simplify the result.

Tap for more steps...

Step 7.2.1

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

$f (1) = 0$

Step 7.2.2

The final answer is $0$ .

$0$

Step 8

Evaluate $f (b)$ from the interval $[1, 8]$ .

Tap for more steps...

Step 8.1

Replace the variable $x$ with $8$ in the expression.

$f (8) = \ln (8)$

Step 8.2

The final answer is $\ln (8)$ .

$\ln (8)$

Step 9

Solve $\frac{1}{x} = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $\frac{1}{x} = \frac{- (f (8) + f (1))}{- (8 + 1)}$ .

Tap for more steps...

Step 9.1

Factor each term.

Tap for more steps...

Step 9.1.1

Multiply $- 1$ by $0$ .

$\frac{1}{x} = \frac{\ln (8) + 0}{(8) - (1)}$

Step 9.1.2

Add $\ln (8)$ and $0$ .

$\frac{1}{x} = \frac{\ln (8)}{(8) - (1)}$

Step 9.1.3

Multiply $- 1$ by $1$ .

$\frac{1}{x} = \frac{\ln (8)}{8 - 1}$

Step 9.1.4

Subtract $1$ from $8$ .

$\frac{1}{x} = \frac{\ln (8)}{7}$

Step 9.1.5

Rewrite $\frac{\ln (8)}{7}$ as $\frac{1}{7} \ln (8)$ .

$\frac{1}{x} = \frac{1}{7} \ln (8)$

Step 9.1.6

Simplify $\frac{1}{7} \ln (8)$ by moving $\frac{1}{7}$ inside the logarithm.

$\frac{1}{x} = \ln (8^{\frac{1}{7}})$

Step 9.2

Find the LCD of the terms in the equation.

Tap for more steps...

Step 9.2.1

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

$x, 1$

Step 9.2.2

The LCM of one and any expression is the expression.

$x$

Step 9.3

Multiply each term in $\frac{1}{x} = \ln (8^{\frac{1}{7}})$ by $x$ to eliminate the fractions.

Tap for more steps...

Step 9.3.1

Multiply each term in $\frac{1}{x} = \ln (8^{\frac{1}{7}})$ by $x$ .

$\frac{1}{x} x = \ln (8^{\frac{1}{7}}) x$

Step 9.3.2

Simplify the left side.

Tap for more steps...

Step 9.3.2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 9.3.2.1.1

Cancel the common factor.

$\frac{1}{x} x = \ln (8^{\frac{1}{7}}) x$

Step 9.3.2.1.2

Rewrite the expression.

$1 = \ln (8^{\frac{1}{7}}) x$

Step 9.3.3

Simplify the right side.

Tap for more steps...

Step 9.3.3.1

Reorder factors in $\ln (8^{\frac{1}{7}}) x$ .

$1 = x \ln (8^{\frac{1}{7}})$

Step 9.4

Solve the equation.

Tap for more steps...

Step 9.4.1

Rewrite the equation as $x \ln (8^{\frac{1}{7}}) = 1$ .

$x \ln (8^{\frac{1}{7}}) = 1$

Step 9.4.2

Divide each term in $x \ln (8^{\frac{1}{7}}) = 1$ by $\ln (8^{\frac{1}{7}})$ and simplify.

Tap for more steps...

Step 9.4.2.1

Divide each term in $x \ln (8^{\frac{1}{7}}) = 1$ by $\ln (8^{\frac{1}{7}})$ .

$\frac{x \ln (8^{\frac{1}{7}})}{\ln (8^{\frac{1}{7}})} = \frac{1}{\ln (8^{\frac{1}{7}})}$

Step 9.4.2.2

Simplify the left side.

Tap for more steps...

Step 9.4.2.2.1

Cancel the common factor.

$\frac{x \ln (8^{\frac{1}{7}})}{\ln (8^{\frac{1}{7}})} = \frac{1}{\ln (8^{\frac{1}{7}})}$

Step 9.4.2.2.2

Divide $x$ by $1$ .

$x = \frac{1}{\ln (8^{\frac{1}{7}})}$

Step 10

There is a tangent line found at $x = \frac{1}{\ln (8^{\frac{1}{7}})}$ parallel to the line that passes through the end points $a = 1$ and $b = 8$ .

There is a tangent line at $x = \frac{1}{\ln (8^{\frac{1}{7}})}$ parallel to the line that passes through the end points $a = 1$ and $b = 8$

Step 11