Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 weeks 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

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  • Alan
    Lv 7
    3 weeks ago
    Best 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 

  • 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

  • 3 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)

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