Find the Maximum/Minimum Value f(x)=2x^2-3x+1

$f (x) = 2 x^{2} - 3 x + 1$

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

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

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

Remove parentheses.

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

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

Multiply $2$ by $2$ .

$x = - \frac{- 3}{4}$

Move the negative in front of the fraction.

$x = - - \frac{3}{4}$

Multiply $- - \frac{3}{4}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{3}{4})$

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

$x = \frac{3}{4}$

Step 3

Evaluate $f (\frac{3}{4})$ .

Replace the variable $x$ with $\frac{3}{4}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{3}{4}$ .

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

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

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

Raise $4$ to the power of $2$ .

$f (\frac{3}{4}) = 2 (\frac{9}{16}) - 3 (\frac{3}{4}) + 1$

Cancel the common factor of $2$ .

Factor $2$ out of $16$ .

$f (\frac{3}{4}) = 2 (\frac{9}{2 (8)}) - 3 (\frac{3}{4}) + 1$

Cancel the common factor.

$f (\frac{3}{4}) = 2 (\frac{9}{2 \cdot 8}) - 3 (\frac{3}{4}) + 1$

Rewrite the expression.

$f (\frac{3}{4}) = \frac{9}{8} - 3 (\frac{3}{4}) + 1$

Multiply $- 3 (\frac{3}{4})$ .

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

$f (\frac{3}{4}) = \frac{9}{8} + \frac{- 3 \cdot 3}{4} + 1$

Multiply $- 3$ by $3$ .

$f (\frac{3}{4}) = \frac{9}{8} + \frac{- 9}{4} + 1$

Move the negative in front of the fraction.

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9}{4} + 1$

Find the common denominator.

Multiply $\frac{9}{4}$ by $\frac{2}{2}$ .

$f (\frac{3}{4}) = \frac{9}{8} - (\frac{9}{4} \cdot \frac{2}{2}) + 1$

Multiply $\frac{9}{4}$ by $\frac{2}{2}$ .

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9 \cdot 2}{4 \cdot 2} + 1$

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

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9 \cdot 2}{4 \cdot 2} + \frac{1}{1}$

Multiply $\frac{1}{1}$ by $\frac{8}{8}$ .

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9 \cdot 2}{4 \cdot 2} + \frac{1}{1} \cdot \frac{8}{8}$

Multiply $\frac{1}{1}$ by $\frac{8}{8}$ .

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9 \cdot 2}{4 \cdot 2} + \frac{8}{8}$

Reorder the factors of $4 \cdot 2$ .

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9 \cdot 2}{2 \cdot 4} + \frac{8}{8}$

Multiply $2$ by $4$ .

$f (\frac{3}{4}) = \frac{9}{8} - \frac{9 \cdot 2}{8} + \frac{8}{8}$

Combine the numerators over the common denominator.

$f (\frac{3}{4}) = \frac{9 - 9 \cdot 2 + 8}{8}$

Simplify the expression.

Multiply $- 9$ by $2$ .

$f (\frac{3}{4}) = \frac{9 - 18 + 8}{8}$

Subtract $18$ from $9$ .

$f (\frac{3}{4}) = \frac{- 9 + 8}{8}$

Add $- 9$ and $8$ .

$f (\frac{3}{4}) = \frac{- 1}{8}$

Move the negative in front of the fraction.

$f (\frac{3}{4}) = - \frac{1}{8}$

The final answer is $- \frac{1}{8}$ .

$- \frac{1}{8}$

Step 4

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

$(\frac{3}{4}, - \frac{1}{8})$

Step 5

Find the Maximum/Minimum Value f(x)=2x^2-3x+1

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