Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 months ago

# 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 Relevance

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

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• 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

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• 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)

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