Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = (x) / (x^2 − x + 25), [0, 15]?

Answer 1

f(x) = (x) / (x^2 − x + 25), for 0 ≤ x ≤ 15

f ' (x) = [(x^2 - x + 25)(1) - x(2x-1) ]/(x^2 - x + 25)^2
= ( x^2 - x + 25 - 2x^2 + x)/(....)^2
= 0 for a max/min

-x^2 + 25 = 0
x = ± 5

if x = 5, f(5) = 5/(25-5+25) = 1/9
if x = -5, f(-5) = -5/(25 + 5 + 25) = -1/11

we should also look at the end valued
f(0) = 0/25 = 0
f(15) = 15/(225 - 15 + 25) = 3/47
for the given interval
max is 3/47
min is -1

local max is 1/9
local min is -1/11