Find the Maximum/Minimum Value f(x)=3x^2-3x

$f (x) = 3 x^{2} - 3 x$

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

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

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

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

Remove parentheses.

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

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

Cancel the common factor of $- 3$ and $3$ .

Factor $3$ out of $- 3$ .

$x = - \frac{3 \cdot - 1}{2 \cdot 3}$

Cancel the common factors.

Factor $3$ out of $2 \cdot 3$ .

$x = - \frac{3 \cdot - 1}{3 \cdot 2}$

Cancel the common factor.

$x = - \frac{3 \cdot - 1}{3 \cdot 2}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

Simplify the result.

Simplify each term.

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

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

One to any power is one.

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

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

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

Combine $3$ and $\frac{1}{4}$ .

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

Combine $- 3$ and $\frac{1}{2}$ .

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

Move the negative in front of the fraction.

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

To write $- \frac{3}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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

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

Multiply $2$ by $2$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 3$ by $2$ .

$f (\frac{1}{2}) = \frac{3 - 6}{4}$

Subtract $6$ from $3$ .

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

Move the negative in front of the fraction.

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

The final answer is $- \frac{3}{4}$ .

$- \frac{3}{4}$

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

$(\frac{1}{2}, - \frac{3}{4})$

Find the Maximum/Minimum Value f(x)=3x^2-3x

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