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

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Answer 1 · 2017-03-01T06:33:48

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

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

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 $x = 5/2$

Find the y-value of the maximum:

$f(5/2) = -(5/2)^2 + 5(5/2) - 1 = -25/4 + 25/2 -1 = -25/4 + 50/4 - 4/4 = 21/4$

relative max at $(5/2, 21/4) = (2.5, 5.25)$

What are the extrema of f(x)=-x^2 +5x -1 f(x)=−x2+5x−1f(x)=-x^2 +5x -1 ?