Calculus Examples

$f (x) = \frac{x}{x + 2}$ , $[1, 4]$

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 $f (x) = \frac{x}{x + 2}$ is continuous.

Tap for more steps...

Step 2.1

To find whether the function is continuous on $[1, 4]$ or not, find the domain of $f (x) = \frac{x}{x + 2}$ .

Tap for more steps...

Step 2.1.1

Set the denominator in $\frac{x}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Step 2.1.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.1.3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2}$

Interval Notation:

$(- \infty, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2}$

Step 2.2

$f (x)$ is continuous on $[1, 4]$ .

The function is continuous.

Step 3

Find the derivative.

Tap for more steps...

Step 3.1

Find the first derivative.

Tap for more steps...

Step 3.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x + 2$ .

$\frac{(x + 2) \frac{d}{d x} [x] - x \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Step 3.1.2

Differentiate.

Tap for more steps...

Step 3.1.2.1

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

$\frac{(x + 2) \cdot 1 - x \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Step 3.1.2.2

Multiply $x + 2$ by $1$ .

$\frac{x + 2 - x \frac{d}{d x} [x + 2]}{{(x + 2)}^{2}}$

Step 3.1.2.3

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$\frac{x + 2 - x (\frac{d}{d x} [x] + \frac{d}{d x} [2])}{{(x + 2)}^{2}}$

Step 3.1.2.4

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

$\frac{x + 2 - x (1 + \frac{d}{d x} [2])}{{(x + 2)}^{2}}$

Step 3.1.2.5

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{x + 2 - x (1 + 0)}{{(x + 2)}^{2}}$

Step 3.1.2.6

Simplify by adding terms.

Tap for more steps...

Step 3.1.2.6.1

Add $1$ and $0$ .

$\frac{x + 2 - x \cdot 1}{{(x + 2)}^{2}}$

Step 3.1.2.6.2

Multiply $- 1$ by $1$ .

$\frac{x + 2 - x}{{(x + 2)}^{2}}$

Step 3.1.2.6.3

Subtract $x$ from $x$ .

$\frac{0 + 2}{{(x + 2)}^{2}}$

Step 3.1.2.6.4

Add $0$ and $2$ .

$f' (x) = \frac{2}{{(x + 2)}^{2}}$

Step 3.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{2}{{(x + 2)}^{2}}$ .

$\frac{2}{{(x + 2)}^{2}}$

Step 4

Find if the derivative is continuous on $(1, 4)$ .

Tap for more steps...

Step 4.1

To find whether the function is continuous on $(1, 4)$ or not, find the domain of $f' (x) = \frac{2}{{(x + 2)}^{2}}$ .

Tap for more steps...

Step 4.1.1

Set the denominator in $\frac{2}{{(x + 2)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 2)}^{2} = 0$

Step 4.1.2

Solve for $x$ .

Tap for more steps...

Step 4.1.2.1

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.1.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.1.3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2}$

Interval Notation:

$(- \infty, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2}$

Step 4.2

$f' (x)$ is continuous on $(1, 4)$ .

The function is continuous.

Step 5

The function is differentiable on $(1, 4)$ because the derivative is continuous on $(1, 4)$ .

The function is differentiable.

Step 6

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[1, 4]$ and differentiable on $(1, 4)$ .

$f (x)$ is continuous on $[1, 4]$ and differentiable on $(1, 4)$ .

Step 7

Evaluate $f (a)$ from the interval $[1, 4]$ .

Tap for more steps...

Step 7.1

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

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

Step 7.2

Simplify the result.

Tap for more steps...

Step 7.2.1

Add $1$ and $2$ .

$f (1) = \frac{1}{3}$

Step 7.2.2

The final answer is $\frac{1}{3}$ .

$\frac{1}{3}$

Step 8

Evaluate $f (b)$ from the interval $[1, 4]$ .

Tap for more steps...

Step 8.1

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

$f (4) = \frac{4}{(4) + 2}$

Step 8.2

Simplify the result.

Tap for more steps...

Step 8.2.1

Cancel the common factor of $4$ and $(4) + 2$ .

Tap for more steps...

Step 8.2.1.1

Factor $2$ out of $4$ .

$f (4) = \frac{2 \cdot 2}{4 + 2}$

Step 8.2.1.2

Cancel the common factors.

Tap for more steps...

Step 8.2.1.2.1

Factor $2$ out of $4$ .

$f (4) = \frac{2 \cdot 2}{2 (2) + 2}$

Step 8.2.1.2.2

Factor $2$ out of $2$ .

$f (4) = \frac{2 \cdot 2}{2 \cdot 2 + 2 \cdot 1}$

Step 8.2.1.2.3

Factor $2$ out of $2 \cdot 2 + 2 \cdot 1$ .

$f (4) = \frac{2 \cdot 2}{2 \cdot (2 + 1)}$

Step 8.2.1.2.4

Cancel the common factor.

$f (4) = \frac{2 \cdot 2}{2 \cdot (2 + 1)}$

Step 8.2.1.2.5

Rewrite the expression.

$f (4) = \frac{2}{2 + 1}$

Step 8.2.2

Add $2$ and $1$ .

$f (4) = \frac{2}{3}$

Step 8.2.3

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

$\frac{2}{3}$

Step 9

Solve $\frac{2}{{(x + 2)}^{2}} = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $\frac{2}{{(x + 2)}^{2}} = \frac{- (f (4) + f (1))}{- (4 + 1)}$ .

Tap for more steps...

Step 9.1

Factor each term.

Tap for more steps...

Step 9.1.1

Multiply $- 1$ by $\frac{1}{3}$ .

$\frac{2}{{(x + 2)}^{2}} = \frac{\frac{2}{3} - \frac{1}{3}}{(4) - (1)}$

Step 9.1.2

Combine the numerators over the common denominator.

$\frac{2}{{(x + 2)}^{2}} = \frac{\frac{2 - 1}{3}}{(4) - (1)}$

Step 9.1.3

Subtract $1$ from $2$ .

$\frac{2}{{(x + 2)}^{2}} = \frac{\frac{1}{3}}{(4) - (1)}$

Step 9.1.4

Multiply $- 1$ by $1$ .

$\frac{2}{{(x + 2)}^{2}} = \frac{\frac{1}{3}}{4 - 1}$

Step 9.1.5

Subtract $1$ from $4$ .

$\frac{2}{{(x + 2)}^{2}} = \frac{\frac{1}{3}}{3}$

Step 9.1.6

Multiply the numerator by the reciprocal of the denominator.

$\frac{2}{{(x + 2)}^{2}} = \frac{1}{3} \cdot \frac{1}{3}$

Step 9.1.7

Multiply $\frac{1}{3} \cdot \frac{1}{3}$ .

Tap for more steps...

Step 9.1.7.1

Multiply $\frac{1}{3}$ by $\frac{1}{3}$ .

$\frac{2}{{(x + 2)}^{2}} = \frac{1}{3 \cdot 3}$

Step 9.1.7.2

Multiply $3$ by $3$ .

$\frac{2}{{(x + 2)}^{2}} = \frac{1}{9}$

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.

${(x + 2)}^{2}, 9$

Step 9.2.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 9.2.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 9.2.4

$9$ has factors of $3$ and $3$ .

$3 \cdot 3$

Step 9.2.5

Multiply $3$ by $3$ .

$9$

Step 9.2.6

The factors for $x + 2$ are $(x + 2) \cdot (x + 2)$ , which is $x + 2$ multiplied by itself $2$ times.

$(x + 2) = (x + 2) \cdot (x + 2)$

$(x + 2)$ occurs $2$ times.

Step 9.2.7

The LCM of ${(x + 2)}^{2}$ is the result of multiplying all factors the greatest number of times they occur in either term.

${(x + 2)}^{2}$

Step 9.2.8

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$9 {(x + 2)}^{2}$

Step 9.3

Multiply each term in $\frac{2}{{(x + 2)}^{2}} = \frac{1}{9}$ by $9 {(x + 2)}^{2}$ to eliminate the fractions.

Tap for more steps...

Step 9.3.1

Multiply each term in $\frac{2}{{(x + 2)}^{2}} = \frac{1}{9}$ by $9 {(x + 2)}^{2}$ .

$\frac{2}{{(x + 2)}^{2}} (9 {(x + 2)}^{2}) = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.2

Simplify the left side.

Tap for more steps...

Step 9.3.2.1

Rewrite using the commutative property of multiplication.

$9 \frac{2}{{(x + 2)}^{2}} {(x + 2)}^{2} = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.2.2

Multiply $9 \frac{2}{{(x + 2)}^{2}}$ .

Tap for more steps...

Step 9.3.2.2.1

Combine $9$ and $\frac{2}{{(x + 2)}^{2}}$ .

$\frac{9 \cdot 2}{{(x + 2)}^{2}} {(x + 2)}^{2} = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.2.2.2

Multiply $9$ by $2$ .

$\frac{18}{{(x + 2)}^{2}} {(x + 2)}^{2} = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.2.3

Cancel the common factor of ${(x + 2)}^{2}$ .

Tap for more steps...

Step 9.3.2.3.1

Cancel the common factor.

$\frac{18}{{(x + 2)}^{2}} {(x + 2)}^{2} = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.2.3.2

Rewrite the expression.

$18 = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.3

Simplify the right side.

Tap for more steps...

Step 9.3.3.1

Cancel the common factor of $9$ .

Tap for more steps...

Step 9.3.3.1.1

Factor $9$ out of $9 {(x + 2)}^{2}$ .

$18 = \frac{1}{9} (9 ({(x + 2)}^{2}))$

Step 9.3.3.1.2

Cancel the common factor.

$18 = \frac{1}{9} (9 {(x + 2)}^{2})$

Step 9.3.3.1.3

Rewrite the expression.

$18 = {(x + 2)}^{2}$

Step 9.4

Solve the equation.

Tap for more steps...

Step 9.4.1

Rewrite the equation as ${(x + 2)}^{2} = 18$ .

${(x + 2)}^{2} = 18$

Step 9.4.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x + 2 = \pm \sqrt{18}$

Step 9.4.3

Simplify $\pm \sqrt{18}$ .

Tap for more steps...

Step 9.4.3.1

Rewrite $18$ as $3^{2} \cdot 2$ .

Tap for more steps...

Step 9.4.3.1.1

Factor $9$ out of $18$ .

$x + 2 = \pm \sqrt{9 (2)}$

Step 9.4.3.1.2

Rewrite $9$ as $3^{2}$ .

$x + 2 = \pm \sqrt{3^{2} \cdot 2}$

Step 9.4.3.2

Pull terms out from under the radical.

$x + 2 = \pm 3 \sqrt{2}$

Step 9.4.4

The complete solution is the result of both the positive and negative portions of the solution.

Tap for more steps...

Step 9.4.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x + 2 = 3 \sqrt{2}$

Step 9.4.4.2

Subtract $2$ from both sides of the equation.

$x = 3 \sqrt{2} - 2$

Step 9.4.4.3

Next, use the negative value of the $\pm$ to find the second solution.

$x + 2 = - 3 \sqrt{2}$

Step 9.4.4.4

Subtract $2$ from both sides of the equation.

$x = - 3 \sqrt{2} - 2$

Step 9.4.4.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 \sqrt{2} - 2, - 3 \sqrt{2} - 2$

Step 10

There is a tangent line found at $x = 3 \sqrt{2} - 2$ parallel to the line that passes through the end points $a = 1$ and $b = 4$ .

There is a tangent line at $x = 3 \sqrt{2} - 2$ parallel to the line that passes through the end points $a = 1$ and $b = 4$

Step 11

There is a tangent line found at $x = - 3 \sqrt{2} - 2$ parallel to the line that passes through the end points $a = 1$ and $b = 4$ .

There is a tangent line at $x = - 3 \sqrt{2} - 2$ parallel to the line that passes through the end points $a = 1$ and $b = 4$

Step 12