Find the Maximum/Minimum Value m(x)=4x-x^2

$m (x) = 4 x - x^{2}$

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 $b$ .

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

Remove parentheses.

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

Simplify $- \frac{4}{2 (- 1)}$ .

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$x = - (- 1 \cdot 2)$

Multiply $- (- 1 \cdot 2)$ .

Multiply $- 1$ by $2$ .

$x = - - 2$

Multiply $- 1$ by $- 2$ .

$x = 2$

Step 3

Evaluate $f (2)$ .

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

$m (2) = 4 (2) - {(2)}^{2}$

Simplify the result.

Simplify each term.

Multiply $4$ by $2$ .

$m (2) = 8 - {(2)}^{2}$

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

$m (2) = 8 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$m (2) = 8 - 4$

Subtract $4$ from $8$ .

$m (2) = 4$

The final answer is $4$ .

$4$

Step 4

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

$(2, 4)$

Step 5

Find the Maximum/Minimum Value m(x)=4x-x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.