Calculus Examples

$f (x) = \frac{x}{x^{2} - x + 25}$ ,

$[0, 15]$

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

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

$g (x) = x^{2} - x + 25$ .

$\frac{(x^{2} - x + 25) \frac{d}{d x} [x] - x \frac{d}{d x} [x^{2} - x + 25]}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$\frac{(x^{2} - x + 25) \cdot 1 - x \frac{d}{d x} [x^{2} - x + 25]}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.2

Multiply

$x^{2} - x + 25$ by

$1$ .

$\frac{x^{2} - x + 25 - x \frac{d}{d x} [x^{2} - x + 25]}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.3

By the Sum Rule, the derivative of

$x^{2} - x + 25$ with respect to

$x$ is

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [25]$ .

$\frac{x^{2} - x + 25 - x (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [25])}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.4

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 2$ .

$\frac{x^{2} - x + 25 - x (2 x + \frac{d}{d x} [- x] + \frac{d}{d x} [25])}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.5

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- x$ with respect to

$x$ is

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

$\frac{x^{2} - x + 25 - x (2 x - \frac{d}{d x} [x] + \frac{d}{d x} [25])}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.6

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$\frac{x^{2} - x + 25 - x (2 x - 1 \cdot 1 + \frac{d}{d x} [25])}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.7

Multiply

$- 1$ by

$1$ .

$\frac{x^{2} - x + 25 - x (2 x - 1 + \frac{d}{d x} [25])}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.8

Since

$25$ is constant with respect to

$x$ , the derivative of

$25$ with respect to

$x$ is

$0$ .

$\frac{x^{2} - x + 25 - x (2 x - 1 + 0)}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.2.9

Add

$2 x - 1$ and

$0$ .

$\frac{x^{2} - x + 25 - x (2 x - 1)}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Apply the distributive property.

$\frac{x^{2} - x + 25 - x (2 x) - x \cdot - 1}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2

Simplify the numerator.

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Step 1.1.1.3.2.1

Simplify each term.

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Step 1.1.1.3.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{x^{2} - x + 25 - 1 \cdot 2 x \cdot x - x \cdot - 1}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.1.2

Multiply

$x$ by

$x$ by adding the exponents.

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Step 1.1.1.3.2.1.2.1

Move

$x$ .

$\frac{x^{2} - x + 25 - 1 \cdot 2 (x \cdot x) - x \cdot - 1}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.1.2.2

Multiply

$x$ by

$x$ .

$\frac{x^{2} - x + 25 - 1 \cdot 2 x^{2} - x \cdot - 1}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.1.3

Multiply

$- 1$ by

$2$ .

$\frac{x^{2} - x + 25 - 2 x^{2} - x \cdot - 1}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.1.4

Multiply

$- x \cdot - 1$ .

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Step 1.1.1.3.2.1.4.1

Multiply

$- 1$ by

$- 1$ .

$\frac{x^{2} - x + 25 - 2 x^{2} + 1 x}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.1.4.2

Multiply

$x$ by

$1$ .

$\frac{x^{2} - x + 25 - 2 x^{2} + x}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.2

Combine the opposite terms in

$x^{2} - x + 25 - 2 x^{2} + x$ .

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Step 1.1.1.3.2.2.1

Add

$- x$ and

$x$ .

$\frac{x^{2} + 25 - 2 x^{2} + 0}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.2.2

Add

$x^{2} + 25 - 2 x^{2}$ and

$0$ .

$\frac{x^{2} + 25 - 2 x^{2}}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.2.3

Subtract

$2 x^{2}$ from

$x^{2}$ .

$\frac{- x^{2} + 25}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.3

Simplify the numerator.

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Step 1.1.1.3.3.1

Rewrite

$25$ as

$5^{2}$ .

$\frac{- x^{2} + 5^{2}}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.3.2

Reorder

$- x^{2}$ and

$5^{2}$ .

$\frac{5^{2} - x^{2}}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.1.3.3.3

Since both terms are perfect squares, factor using the difference of squares formula,

$a^{2} - b^{2} = (a + b) (a - b)$ where

$a = 5$ and

$b = x$ .

$f' (x) = \frac{(5 + x) (5 - x)}{{(x^{2} - x + 25)}^{2}}$

Step 1.1.2

The first derivative of

$f (x)$ with respect to

$x$ is

$\frac{(5 + x) (5 - x)}{{(x^{2} - x + 25)}^{2}}$ .

$\frac{(5 + x) (5 - x)}{{(x^{2} - x + 25)}^{2}}$

Step 1.2

Set the first derivative equal to

$0$ then solve the equation

$\frac{(5 + x) (5 - x)}{{(x^{2} - x + 25)}^{2}} = 0$ .

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Step 1.2.1

Set the first derivative equal to

$0$ .

$\frac{(5 + x) (5 - x)}{{(x^{2} - x + 25)}^{2}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$(5 + x) (5 - x) = 0$

Step 1.2.3

Solve the equation for

$x$ .

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Step 1.2.3.1

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$5 + x = 0$

$5 - x = 0$

Step 1.2.3.2

Set

$5 + x$ equal to

$0$ and solve for

$x$ .

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Step 1.2.3.2.1

Set

$5 + x$ equal to

$0$ .

$5 + x = 0$

Step 1.2.3.2.2

Subtract

$5$ from both sides of the equation.

$x = - 5$

Step 1.2.3.3

Set

$5 - x$ equal to

$0$ and solve for

$x$ .

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Step 1.2.3.3.1

Set

$5 - x$ equal to

$0$ .

$5 - x = 0$

Step 1.2.3.3.2

Solve

$5 - x = 0$ for

$x$ .

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Step 1.2.3.3.2.1

Subtract

$5$ from both sides of the equation.

$- x = - 5$

Step 1.2.3.3.2.2

Divide each term in

$- x = - 5$ by

$- 1$ and simplify.

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Step 1.2.3.3.2.2.1

Divide each term in

$- x = - 5$ by

$- 1$ .

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

Step 1.2.3.3.2.2.2

Simplify the left side.

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Step 1.2.3.3.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 1.2.3.3.2.2.2.2

Divide

$x$ by

$1$ .

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

Step 1.2.3.3.2.2.3

Simplify the right side.

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Step 1.2.3.3.2.2.3.1

Divide

$- 5$ by

$- 1$ .

$x = 5$

Step 1.2.3.4

The final solution is all the values that make

$(5 + x) (5 - x) = 0$ true.

$x = - 5, 5$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 1.4

Evaluate

$\frac{x}{x^{2} - x + 25}$ at each

$x$ value where the derivative is

$0$ or undefined.

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Step 1.4.1

Evaluate at

$x = - 5$ .

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Step 1.4.1.1

Substitute

$- 5$ for

$x$ .

$\frac{- 5}{{(- 5)}^{2} - (- 5) + 25}$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify the denominator.

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Step 1.4.1.2.1.1

Raise

$- 5$ to the power of

$2$ .

$\frac{- 5}{25 - (- 5) + 25}$

Step 1.4.1.2.1.2

Multiply

$- 1$ by

$- 5$ .

$\frac{- 5}{25 + 5 + 25}$

Step 1.4.1.2.1.3

Add

$25$ and

$5$ .

$\frac{- 5}{30 + 25}$

Step 1.4.1.2.1.4

Add

$30$ and

$25$ .

$\frac{- 5}{55}$

Step 1.4.1.2.2

Reduce the expression by cancelling the common factors.

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Step 1.4.1.2.2.1

Cancel the common factor of

$- 5$ and

$55$ .

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Step 1.4.1.2.2.1.1

Factor

$5$ out of

$- 5$ .

$\frac{5 (- 1)}{55}$

Step 1.4.1.2.2.1.2

Cancel the common factors.

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Step 1.4.1.2.2.1.2.1

Factor

$5$ out of

$55$ .

$\frac{5 \cdot - 1}{5 \cdot 11}$

Step 1.4.1.2.2.1.2.2

Cancel the common factor.

$\frac{5 \cdot - 1}{5 \cdot 11}$

Step 1.4.1.2.2.1.2.3

Rewrite the expression.

$\frac{- 1}{11}$

Step 1.4.1.2.2.2

Move the negative in front of the fraction.

$- \frac{1}{11}$

Step 1.4.2

Evaluate at

$x = 5$ .

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Step 1.4.2.1

Substitute

$5$ for

$x$ .

$\frac{5}{{(5)}^{2} - (5) + 25}$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

Cancel the common factor of

$5$ and

${(5)}^{2} - (5) + 25$ .

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Step 1.4.2.2.1.1

Factor

$5$ out of

$5$ .

$\frac{5 \cdot 1}{5^{2} - (5) + 25}$

Step 1.4.2.2.1.2

Cancel the common factors.

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Step 1.4.2.2.1.2.1

Factor

$5$ out of

$5^{2}$ .

$\frac{5 \cdot 1}{5 \cdot 5 - (5) + 25}$

Step 1.4.2.2.1.2.2

Factor

$5$ out of

$- (5)$ .

$\frac{5 \cdot 1}{5 \cdot 5 + 5 \cdot - 1 + 25}$

Step 1.4.2.2.1.2.3

Factor

$5$ out of

$5 \cdot 5 + 5 \cdot - 1$ .

$\frac{5 \cdot 1}{5 \cdot (5 - 1) + 25}$

Step 1.4.2.2.1.2.4

Factor

$5$ out of

$25$ .

$\frac{5 \cdot 1}{5 \cdot (5 - 1) + 5 (5)}$

Step 1.4.2.2.1.2.5

Factor

$5$ out of

$5 \cdot (5 - 1) + 5 (5)$ .

$\frac{5 \cdot 1}{5 \cdot (5 - 1 + 5)}$

Step 1.4.2.2.1.2.6

Cancel the common factor.

$\frac{5 \cdot 1}{5 \cdot (5 - 1 + 5)}$

Step 1.4.2.2.1.2.7

Rewrite the expression.

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

Step 1.4.2.2.2

Simplify the denominator.

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Step 1.4.2.2.2.1

Subtract

$1$ from

$5$ .

$\frac{1}{4 + 5}$

Step 1.4.2.2.2.2

Add

$4$ and

$5$ .

$\frac{1}{9}$

Step 1.4.3

List all of the points.

$(- 5, - \frac{1}{11}), (5, \frac{1}{9})$

Step 2

Exclude the points that are not on the interval.

$(5, \frac{1}{9})$

Step 3

Evaluate at the included endpoints.

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Step 3.1

Evaluate at

$x = 0$ .

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Step 3.1.1

Substitute

$0$ for

$x$ .

$\frac{0}{{(0)}^{2} - (0) + 25}$

Step 3.1.2

Simplify.

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Step 3.1.2.1

Simplify the denominator.

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Step 3.1.2.1.1

Raising

$0$ to any positive power yields

$0$ .

$\frac{0}{0 - (0) + 25}$

Step 3.1.2.1.2

Multiply

$- 1$ by

$0$ .

$\frac{0}{0 + 0 + 25}$

Step 3.1.2.1.3

Add

$0$ and

$0$ .

$\frac{0}{0 + 25}$

Step 3.1.2.1.4

Add

$0$ and

$25$ .

$\frac{0}{25}$

Step 3.1.2.2

Divide

$0$ by

$25$ .

$0$

Step 3.2

Evaluate at

$x = 15$ .

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Step 3.2.1

Substitute

$15$ for

$x$ .

$\frac{15}{{(15)}^{2} - (15) + 25}$

Step 3.2.2

Simplify.

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Step 3.2.2.1

Simplify the denominator.

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Step 3.2.2.1.1

Raise

$15$ to the power of

$2$ .

$\frac{15}{225 - (15) + 25}$

Step 3.2.2.1.2

Multiply

$- 1$ by

$15$ .

$\frac{15}{225 - 15 + 25}$

Step 3.2.2.1.3

Subtract

$15$ from

$225$ .

$\frac{15}{210 + 25}$

Step 3.2.2.1.4

Add

$210$ and

$25$ .

$\frac{15}{235}$

Step 3.2.2.2

Cancel the common factor of

$15$ and

$235$ .

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Step 3.2.2.2.1

Factor

$5$ out of

$15$ .

$\frac{5 (3)}{235}$

Step 3.2.2.2.2

Cancel the common factors.

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Step 3.2.2.2.2.1

Factor

$5$ out of

$235$ .

$\frac{5 \cdot 3}{5 \cdot 47}$

Step 3.2.2.2.2.2

Cancel the common factor.

$\frac{5 \cdot 3}{5 \cdot 47}$

Step 3.2.2.2.2.3

Rewrite the expression.

$\frac{3}{47}$

Step 3.3

List all of the points.

$(0, 0), (15, \frac{3}{47})$

Step 4

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:

$(5, \frac{1}{9})$

Absolute Minimum:

$(0, 0)$

Step 5