Find the Maximum/Minimum Value 3+3x-x^2

$3 + 3 x - x^{2}$

Step 1

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

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

Step 2

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

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

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

Remove parentheses.

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

Simplify $- \frac{3}{2 (- 1)}$ .

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{3}{2}$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{3}{2}$ by $1$ .

$x = \frac{3}{2}$

Step 3

Evaluate $f (\frac{3}{2})$ .

Replace the variable $x$ with $\frac{3}{2}$ in the expression.

$f (\frac{3}{2}) = 3 + 3 (\frac{3}{2}) - {(\frac{3}{2})}^{2}$

Simplify the result.

Simplify each term.

Multiply $3 (\frac{3}{2})$ .

Combine $3$ and $\frac{3}{2}$ .

$f (\frac{3}{2}) = 3 + \frac{3 \cdot 3}{2} - {(\frac{3}{2})}^{2}$

Multiply $3$ by $3$ .

$f (\frac{3}{2}) = 3 + \frac{9}{2} - {(\frac{3}{2})}^{2}$

Apply the product rule to $\frac{3}{2}$ .

$f (\frac{3}{2}) = 3 + \frac{9}{2} - \frac{3^{2}}{2^{2}}$

Raise $3$ to the power of $2$ .

$f (\frac{3}{2}) = 3 + \frac{9}{2} - \frac{9}{2^{2}}$

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

$f (\frac{3}{2}) = 3 + \frac{9}{2} - \frac{9}{4}$

Find the common denominator.

Write $3$ as a fraction with denominator $1$ .

$f (\frac{3}{2}) = \frac{3}{1} + \frac{9}{2} - \frac{9}{4}$

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$f (\frac{3}{2}) = \frac{3}{1} \cdot \frac{4}{4} + \frac{9}{2} - \frac{9}{4}$

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$f (\frac{3}{2}) = \frac{3 \cdot 4}{4} + \frac{9}{2} - \frac{9}{4}$

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (\frac{3}{2}) = \frac{3 \cdot 4}{4} + \frac{9}{2} \cdot \frac{2}{2} - \frac{9}{4}$

Multiply $\frac{9}{2}$ by $\frac{2}{2}$ .

$f (\frac{3}{2}) = \frac{3 \cdot 4}{4} + \frac{9 \cdot 2}{2 \cdot 2} - \frac{9}{4}$

Multiply $2$ by $2$ .

$f (\frac{3}{2}) = \frac{3 \cdot 4}{4} + \frac{9 \cdot 2}{4} - \frac{9}{4}$

Combine the numerators over the common denominator.

$f (\frac{3}{2}) = \frac{3 \cdot 4 + 9 \cdot 2 - 9}{4}$

Simplify each term.

Multiply $3$ by $4$ .

$f (\frac{3}{2}) = \frac{12 + 9 \cdot 2 - 9}{4}$

Multiply $9$ by $2$ .

$f (\frac{3}{2}) = \frac{12 + 18 - 9}{4}$

Simplify by adding and subtracting.

Add $12$ and $18$ .

$f (\frac{3}{2}) = \frac{30 - 9}{4}$

Subtract $9$ from $30$ .

$f (\frac{3}{2}) = \frac{21}{4}$

The final answer is $\frac{21}{4}$ .

$\frac{21}{4}$

Step 4

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

$(\frac{3}{2}, \frac{21}{4})$

Step 5

Find the Maximum/Minimum Value 3+3x-x^2

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