# Find the absolute maximum and absolute minimum values of the function 𝑓(𝑥)=𝑥^2+(2/𝑥) on the interval [2.5,6].?

It's not 6,2.5 or 4,1

### 3 Answers

- AlanLv 73 weeks agoFavorite Answer
f'(x) = 2x - (2/x^2)

so a turning point would be where

2x- (2/x^2) = 0

2x = 2/x^2

2x^3 = 2

x^3 = 1

x = 1

but one is outside of your range, so you can ignore it.

so between 2.5 and 6 f'(x) is positive so it is increasing over the whole interval

so the minimum and maximum will be at the endpoints

f(2.5) = (2.5)^2 + (2/2.5) = 6.25 + 0.8= 7.05

f(6) = (6)^2 + 2/6 = 36 + 1/3 = 36 1/3

so they are asking for the maximum and minimum values and

not the coordinate pair of the minimum and maximum

absolute min = 7.05

so absolute max = 36 1/3 = (approx.) 36.333333

- 3 weeks ago
You can always graph it on desmos

What we do is take the first derivative and set it equal to 0. That yields the y value and then we plug that into the original function for the x value. I think that's how it goes

- L. E. GantLv 73 weeks ago
f'(x) = 2x - 2/x^2

local max or min when f'(x) = 0 and f"(x) not = 0

f"(x) = 2 + 4/x^3 2x - 2/x^2 = 0 ==> x - 1/x^2 = 0==> x^3 - 1 = 0--> x^3 = 1==> x = 1, outside the range (2.5, 6)In the range, f"(x) > 0 -- always increasing so f2.5) = 7.05 (absolute minimum in the range)f(6) =109/3 (absolute maximum in the range)