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

Question

2335 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T06:33:48

relative max at #(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)#