Calculus Examples

$y = x^{2} + x + 2$

Step 1

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

Find the value of $x = - \frac{b}{2 a}$ .

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Step 2.1

Substitute in the values of $a$ and $b$ .

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

Step 2.2

Remove parentheses.

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

Step 2.3

Cancel the common factor of $1$ .

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Step 2.3.1

Cancel the common factor.

$x = - \frac{1}{2 \cdot 1}$

Step 2.3.2

Rewrite the expression.

$x = - \frac{1}{2}$

Step 3

Evaluate $f (- \frac{1}{2})$ .

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Step 3.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

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

Step 3.2

Simplify the result.

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Step 3.2.1

Remove parentheses.

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

Step 3.2.2

Simplify each term.

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Step 3.2.2.1

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

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Step 3.2.2.1.1

Apply the product rule to $- \frac{1}{2}$ .

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

Step 3.2.2.1.2

Apply the product rule to $\frac{1}{2}$ .

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

Step 3.2.2.2

Raise $- 1$ to the power of $2$ .

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

Step 3.2.2.3

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

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

Step 3.2.2.4

One to any power is one.

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

Step 3.2.2.5

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

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

Step 3.2.3

Find the common denominator.

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Step 3.2.3.1

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

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

Step 3.2.3.2

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

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

Step 3.2.3.3

Write $2$ as a fraction with denominator $1$ .

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

Step 3.2.3.4

Multiply $\frac{2}{1}$ by $\frac{4}{4}$ .

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

Step 3.2.3.5

Multiply $\frac{2}{1}$ by $\frac{4}{4}$ .

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

Step 3.2.3.6

Multiply $2$ by $2$ .

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

Step 3.2.4

Combine the numerators over the common denominator.

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

Step 3.2.5

Simplify the expression.

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Step 3.2.5.1

Multiply $2$ by $4$ .

$f (- \frac{1}{2}) = \frac{1 - 2 + 8}{4}$

Step 3.2.5.2

Subtract $2$ from $1$ .

$f (- \frac{1}{2}) = \frac{- 1 + 8}{4}$

Step 3.2.5.3

Add $- 1$ and $8$ .

$f (- \frac{1}{2}) = \frac{7}{4}$

Step 3.2.6

The final answer is $\frac{7}{4}$ .

$\frac{7}{4}$

Step 4

Use the $x$ and $y$ values to find where the minimum occurs.

$(- \frac{1}{2}, \frac{7}{4})$

Step 5