f' = 0 = 4 - 2 x
so x = 2, right
at 2
y = 17 + 8 - 4 = 21
check end points
at x = 0, y = 17
at x = 5, y = 17+20 -25 =12
21 is at the vertex and is maximum
12 is minimum on this restricted domain
Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = 17 + 4x − x^2, [0, 5]
I got -13 and 17, but the answers were wrong. Take the derivative for the critical values, right? To get 2 as a critical value. What from there?
1 answer