# What are the extrema of f(x)=-x^2 +5x -1 ?

Mar 1, 2017

relative max at $\left(\frac{5}{2} , \frac{21}{4}\right) = \left(2.5 , 5.25\right)$

#### Explanation:

Find the first derivative: $f \left(x\right) ' = - 2 x + 5$

Find the critical number(s): f'(x) = 0; x = 5/2

Use the 2nd derivative test to see if the critical number is a relative max. or relative min.:

f''(x) = -2; f''(5/2) < 0; relative max. at $x = \frac{5}{2}$

Find the y-value of the maximum:

$f \left(\frac{5}{2}\right) = - {\left(\frac{5}{2}\right)}^{2} + 5 \left(\frac{5}{2}\right) - 1 = - \frac{25}{4} + \frac{25}{2} - 1 = - \frac{25}{4} + \frac{50}{4} - \frac{4}{4} = \frac{21}{4}$

relative max at $\left(\frac{5}{2} , \frac{21}{4}\right) = \left(2.5 , 5.25\right)$