Calculus Examples

f (x) = 10 x^{2} + x

$f (x) = 10 x^{2} + x$

Step 1

The minimum of a quadratic function occurs at

x = - \frac{b}{2 a}

$x = - \frac{b}{2 a}$ . If

a

$a$ is positive, the minimum value of the function is

f (- \frac{b}{2 a})

$f (- \frac{b}{2 a})$ .

f_{min}

$f_{min}$

x = a x^{2} + b x + c

$x = a x^{2} + b x + c$ occurs at

x = - \frac{b}{2 a}

$x = - \frac{b}{2 a}$

Step 2

Find the value of

x = - \frac{b}{2 a}

$x = - \frac{b}{2 a}$ .

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Step 2.1

Substitute in the values of

a

$a$ and

b

$b$ .

x = - \frac{1}{2 (10)}

$x = - \frac{1}{2 (10)}$

Step 2.2

Remove parentheses.

x = - \frac{1}{2 (10)}

$x = - \frac{1}{2 (10)}$

Step 2.3

Multiply

2

$2$ by

10

$10$ .

x = - \frac{1}{20}

$x = - \frac{1}{20}$

x = - \frac{1}{20}

$x = - \frac{1}{20}$

Step 3

Evaluate

f (- \frac{1}{20})

$f (- \frac{1}{20})$ .

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Step 3.1

Replace the variable

x

$x$ with

- \frac{1}{20}

$- \frac{1}{20}$ in the expression.

f (- \frac{1}{20}) = 10 {(- \frac{1}{20})}^{2} - \frac{1}{20}

$f (- \frac{1}{20}) = 10 {(- \frac{1}{20})}^{2} - \frac{1}{20}$

Step 3.2

Simplify the result.

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Step 3.2.1

Remove parentheses.

f (- \frac{1}{20}) = 10 {(- \frac{1}{20})}^{2} - \frac{1}{20}

$f (- \frac{1}{20}) = 10 {(- \frac{1}{20})}^{2} - \frac{1}{20}$

Step 3.2.2

Simplify each term.

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Step 3.2.2.1

Use the power rule

{(a b)}^{n} = a^{n} b^{n}

${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.2.2.1.1

Apply the product rule to

- \frac{1}{20}

$- \frac{1}{20}$ .

f (- \frac{1}{20}) = 10 ({(- 1)}^{2} {(\frac{1}{20})}^{2}) - \frac{1}{20}

$f (- \frac{1}{20}) = 10 ({(- 1)}^{2} {(\frac{1}{20})}^{2}) - \frac{1}{20}$

Step 3.2.2.1.2

Apply the product rule to

\frac{1}{20}

$\frac{1}{20}$ .

f (- \frac{1}{20}) = 10 ({(- 1)}^{2} (\frac{1^{2}}{20^{2}})) - \frac{1}{20}

$f (- \frac{1}{20}) = 10 ({(- 1)}^{2} (\frac{1^{2}}{20^{2}})) - \frac{1}{20}$

f (- \frac{1}{20}) = 10 ({(- 1)}^{2} (\frac{1^{2}}{20^{2}})) - \frac{1}{20}

$f (- \frac{1}{20}) = 10 ({(- 1)}^{2} (\frac{1^{2}}{20^{2}})) - \frac{1}{20}$

Step 3.2.2.2

Raise

- 1

$- 1$ to the power of

2

$2$ .

f (- \frac{1}{20}) = 10 (1 (\frac{1^{2}}{20^{2}})) - \frac{1}{20}

$f (- \frac{1}{20}) = 10 (1 (\frac{1^{2}}{20^{2}})) - \frac{1}{20}$

Step 3.2.2.3

Multiply

$\frac{1^{2}}{20^{2}}$ by

$1$ .

$f (- \frac{1}{20}) = 10 (\frac{1^{2}}{20^{2}}) - \frac{1}{20}$

Step 3.2.2.4

One to any power is one.

$f (- \frac{1}{20}) = 10 (\frac{1}{20^{2}}) - \frac{1}{20}$

Step 3.2.2.5

Raise

$20$ to the power of

$2$ .

$f (- \frac{1}{20}) = 10 (\frac{1}{400}) - \frac{1}{20}$

Step 3.2.2.6

Cancel the common factor of

$10$ .

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Step 3.2.2.6.1

Factor

$10$ out of

$400$ .

$f (- \frac{1}{20}) = 10 (\frac{1}{10 (40)}) - \frac{1}{20}$

Step 3.2.2.6.2

Cancel the common factor.

$f (- \frac{1}{20}) = 10 (\frac{1}{10 \cdot 40}) - \frac{1}{20}$

Step 3.2.2.6.3

Rewrite the expression.

$f (- \frac{1}{20}) = \frac{1}{40} - \frac{1}{20}$

Step 3.2.3

To write

$- \frac{1}{20}$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

$f (- \frac{1}{20}) = \frac{1}{40} - \frac{1}{20} \cdot \frac{2}{2}$

Step 3.2.4

Write each expression with a common denominator of

$40$ , by multiplying each by an appropriate factor of

$1$ .

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Step 3.2.4.1

Multiply

$\frac{1}{20}$ by

$\frac{2}{2}$ .

$f (- \frac{1}{20}) = \frac{1}{40} - \frac{2}{20 \cdot 2}$

Step 3.2.4.2

Multiply

$20$ by

$2$ .

$f (- \frac{1}{20}) = \frac{1}{40} - \frac{2}{40}$

Step 3.2.5

Combine the numerators over the common denominator.

$f (- \frac{1}{20}) = \frac{1 - 2}{40}$

Step 3.2.6

Subtract

$2$ from

$1$ .

$f (- \frac{1}{20}) = \frac{- 1}{40}$

Step 3.2.7

Move the negative in front of the fraction.

$f (- \frac{1}{20}) = - \frac{1}{40}$

Step 3.2.8

The final answer is

$- \frac{1}{40}$ .

$- \frac{1}{40}$

Step 4

Use the

$x$ and

$y$ values to find where the minimum occurs.

$(- \frac{1}{20}, - \frac{1}{40})$

Step 5