Calculus Examples

$f (x) = \frac{x^{2}}{x^{2} + 6}$ , $[- 2, 2]$

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 $f (x) = \frac{x^{2}}{x^{2} + 6}$ is continuous.

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Step 2.1

To find whether the function is continuous on $[- 2, 2]$ or not, find the domain of $f (x) = \frac{x^{2}}{x^{2} + 6}$ .

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Step 2.1.1

Set the denominator in $\frac{x^{2}}{x^{2} + 6}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 6 = 0$

Step 2.1.2

Solve for $x$ .

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Step 2.1.2.1

Subtract $6$ from both sides of the equation.

$x^{2} = - 6$

Step 2.1.2.2

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

$x = \pm \sqrt{- 6}$

Step 2.1.2.3

Simplify $\pm \sqrt{- 6}$ .

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Step 2.1.2.3.1

Rewrite $- 6$ as $- 1 (6)$ .

$x = \pm \sqrt{- 1 (6)}$

Step 2.1.2.3.2

Rewrite $\sqrt{- 1 (6)}$ as $\sqrt{- 1} \cdot \sqrt{6}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{6}$

Step 2.1.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{6}$

Step 2.1.2.4

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

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Step 2.1.2.4.1

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

$x = i \sqrt{6}$

Step 2.1.2.4.2

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

$x = - i \sqrt{6}$

Step 2.1.2.4.3

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

$x = i \sqrt{6}, - i \sqrt{6}$

Step 2.1.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 2.2

$f (x)$ is continuous on $[- 2, 2]$ .

The function is continuous.

Step 3

Find the derivative.

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Step 3.1

Find the first derivative.

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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 $f (x) = x^{2}$ and $g (x) = x^{2} + 6$ .

$\frac{(x^{2} + 6) \frac{d}{d x} [x^{2}] - x^{2} \frac{d}{d x} [x^{2} + 6]}{{(x^{2} + 6)}^{2}}$

Step 3.1.2

Differentiate.

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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 $n = 2$ .

$\frac{(x^{2} + 6) (2 x) - x^{2} \frac{d}{d x} [x^{2} + 6]}{{(x^{2} + 6)}^{2}}$

Step 3.1.2.2

Move $2$ to the left of $x^{2} + 6$ .

$\frac{2 \cdot (x^{2} + 6) x - x^{2} \frac{d}{d x} [x^{2} + 6]}{{(x^{2} + 6)}^{2}}$

Step 3.1.2.3

By the Sum Rule, the derivative of $x^{2} + 6$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$ .

$\frac{2 (x^{2} + 6) x - x^{2} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [6])}{{(x^{2} + 6)}^{2}}$

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 $n = 2$ .

$\frac{2 (x^{2} + 6) x - x^{2} (2 x + \frac{d}{d x} [6])}{{(x^{2} + 6)}^{2}}$

Step 3.1.2.5

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$\frac{2 (x^{2} + 6) x - x^{2} (2 x + 0)}{{(x^{2} + 6)}^{2}}$

Step 3.1.2.6

Simplify the expression.

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Step 3.1.2.6.1

Add $2 x$ and $0$ .

$\frac{2 (x^{2} + 6) x - x^{2} (2 x)}{{(x^{2} + 6)}^{2}}$

Step 3.1.2.6.2

Multiply $2$ by $- 1$ .

$\frac{2 (x^{2} + 6) x - 2 x^{2} x}{{(x^{2} + 6)}^{2}}$

Step 3.1.3

Raise $x$ to the power of $1$ .

$\frac{2 (x^{2} + 6) x - 2 (x^{1} x^{2})}{{(x^{2} + 6)}^{2}}$

Step 3.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 (x^{2} + 6) x - 2 x^{1 + 2}}{{(x^{2} + 6)}^{2}}$

Step 3.1.5

Add $1$ and $2$ .

$\frac{2 (x^{2} + 6) x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6

Simplify.

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Step 3.1.6.1

Apply the distributive property.

$\frac{(2 x^{2} + 2 \cdot 6) x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.2

Apply the distributive property.

$\frac{2 x^{2} x + 2 \cdot 6 x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3

Simplify the numerator.

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Step 3.1.6.3.1

Simplify each term.

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Step 3.1.6.3.1.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 3.1.6.3.1.1.1

Move $x$ .

$\frac{2 (x \cdot x^{2}) + 2 \cdot 6 x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3.1.1.2

Multiply $x$ by $x^{2}$ .

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Step 3.1.6.3.1.1.2.1

Raise $x$ to the power of $1$ .

$\frac{2 (x^{1} x^{2}) + 2 \cdot 6 x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3.1.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 x^{1 + 2} + 2 \cdot 6 x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3.1.1.3

Add $1$ and $2$ .

$\frac{2 x^{3} + 2 \cdot 6 x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3.1.2

Multiply $2$ by $6$ .

$\frac{2 x^{3} + 12 x - 2 x^{3}}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3.2

Combine the opposite terms in $2 x^{3} + 12 x - 2 x^{3}$ .

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Step 3.1.6.3.2.1

Subtract $2 x^{3}$ from $2 x^{3}$ .

$\frac{12 x + 0}{{(x^{2} + 6)}^{2}}$

Step 3.1.6.3.2.2

Add $12 x$ and $0$ .

$f' (x) = \frac{12 x}{{(x^{2} + 6)}^{2}}$

Step 3.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{12 x}{{(x^{2} + 6)}^{2}}$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}}$

Step 4

Find if the derivative is continuous on $(- 2, 2)$ .

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Step 4.1

To find whether the function is continuous on $(- 2, 2)$ or not, find the domain of $f' (x) = \frac{12 x}{{(x^{2} + 6)}^{2}}$ .

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Step 4.1.1

Set the denominator in $\frac{12 x}{{(x^{2} + 6)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x^{2} + 6)}^{2} = 0$

Step 4.1.2

Solve for $x$ .

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Step 4.1.2.1

Set the $x^{2} + 6$ equal to $0$ .

$x^{2} + 6 = 0$

Step 4.1.2.2

Solve for $x$ .

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Step 4.1.2.2.1

Subtract $6$ from both sides of the equation.

$x^{2} = - 6$

Step 4.1.2.2.2

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

$x = \pm \sqrt{- 6}$

Step 4.1.2.2.3

Simplify $\pm \sqrt{- 6}$ .

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Step 4.1.2.2.3.1

Rewrite $- 6$ as $- 1 (6)$ .

$x = \pm \sqrt{- 1 (6)}$

Step 4.1.2.2.3.2

Rewrite $\sqrt{- 1 (6)}$ as $\sqrt{- 1} \cdot \sqrt{6}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{6}$

Step 4.1.2.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{6}$

Step 4.1.2.2.4

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

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Step 4.1.2.2.4.1

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

$x = i \sqrt{6}$

Step 4.1.2.2.4.2

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

$x = - i \sqrt{6}$

Step 4.1.2.2.4.3

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

$x = i \sqrt{6}, - i \sqrt{6}$

Step 4.1.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4.2

$f' (x)$ is continuous on $(- 2, 2)$ .

The function is continuous.

Step 5

The function is differentiable on $(- 2, 2)$ because the derivative is continuous on $(- 2, 2)$ .

The function is differentiable.

Step 6

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[- 2, 2]$ and differentiable on $(- 2, 2)$ .

$f (x)$ is continuous on $[- 2, 2]$ and differentiable on $(- 2, 2)$ .

Step 7

Evaluate $f (a)$ from the interval $[- 2, 2]$ .

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Step 7.1

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

$f (- 2) = \frac{{(- 2)}^{2}}{{(- 2)}^{2} + 6}$

Step 7.2

Simplify the result.

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Step 7.2.1

Raise $- 2$ to the power of $2$ .

$f (- 2) = \frac{4}{{(- 2)}^{2} + 6}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Raise $- 2$ to the power of $2$ .

$f (- 2) = \frac{4}{4 + 6}$

Step 7.2.2.2

Add $4$ and $6$ .

$f (- 2) = \frac{4}{10}$

Step 7.2.3

Cancel the common factor of $4$ and $10$ .

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Step 7.2.3.1

Factor $2$ out of $4$ .

$f (- 2) = \frac{2 (2)}{10}$

Step 7.2.3.2

Cancel the common factors.

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Step 7.2.3.2.1

Factor $2$ out of $10$ .

$f (- 2) = \frac{2 \cdot 2}{2 \cdot 5}$

Step 7.2.3.2.2

Cancel the common factor.

$f (- 2) = \frac{2 \cdot 2}{2 \cdot 5}$

Step 7.2.3.2.3

Rewrite the expression.

$f (- 2) = \frac{2}{5}$

Step 7.2.4

The final answer is $\frac{2}{5}$ .

$\frac{2}{5}$

Step 8

Solve $\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{- (f (2) + f (- 2))}{- (2 - 2)}$ .

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Step 8.1

Factor each term.

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Step 8.1.1

Multiply $- 1$ by $\frac{2}{5}$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{\frac{2}{5} - \frac{2}{5}}{(2) - (- 2)}$

Step 8.1.2

Combine the numerators over the common denominator.

$\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{\frac{2 - 2}{5}}{(2) - (- 2)}$

Step 8.1.3

Subtract $2$ from $2$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{\frac{0}{5}}{(2) - (- 2)}$

Step 8.1.4

Divide $0$ by $5$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{0}{(2) - (- 2)}$

Step 8.1.5

Multiply $- 1$ by $- 2$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{0}{2 + 2}$

Step 8.1.6

Add $2$ and $2$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} = \frac{0}{4}$

Step 8.1.7

Divide $0$ by $4$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} = 0$

Step 8.2

Find the LCD of the terms in the equation.

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Step 8.2.1

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

${(x^{2} + 6)}^{2}, 1$

Step 8.2.2

The LCM of one and any expression is the expression.

${(x^{2} + 6)}^{2}$

Step 8.3

Multiply each term in $\frac{12 x}{{(x^{2} + 6)}^{2}} = 0$ by ${(x^{2} + 6)}^{2}$ to eliminate the fractions.

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Step 8.3.1

Multiply each term in $\frac{12 x}{{(x^{2} + 6)}^{2}} = 0$ by ${(x^{2} + 6)}^{2}$ .

$\frac{12 x}{{(x^{2} + 6)}^{2}} {(x^{2} + 6)}^{2} = 0 {(x^{2} + 6)}^{2}$

Step 8.3.2

Simplify the left side.

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Step 8.3.2.1

Cancel the common factor of ${(x^{2} + 6)}^{2}$ .

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Step 8.3.2.1.1

Cancel the common factor.

$\frac{12 x}{{(x^{2} + 6)}^{2}} {(x^{2} + 6)}^{2} = 0 {(x^{2} + 6)}^{2}$

Step 8.3.2.1.2

Rewrite the expression.

$12 x = 0 {(x^{2} + 6)}^{2}$

Step 8.3.3

Simplify the right side.

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Step 8.3.3.1

Multiply $0$ by ${(x^{2} + 6)}^{2}$ .

$12 x = 0$

Step 8.4

Divide each term in $12 x = 0$ by $12$ and simplify.

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Step 8.4.1

Divide each term in $12 x = 0$ by $12$ .

$\frac{12 x}{12} = \frac{0}{12}$

Step 8.4.2

Simplify the left side.

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Step 8.4.2.1

Cancel the common factor of $12$ .

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Step 8.4.2.1.1

Cancel the common factor.

$\frac{12 x}{12} = \frac{0}{12}$

Step 8.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{12}$

Step 8.4.3

Simplify the right side.

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Step 8.4.3.1

Divide $0$ by $12$ .

$x = 0$

Step 9

There is a tangent line found at $x = 0$ parallel to the line that passes through the end points $a = - 2$ and $b = 2$ .

There is a tangent line at $x = 0$ parallel to the line that passes through the end points $a = - 2$ and $b = 2$

Step 10