There is a function defined by f:[-1/2,infinity), f(x) = 3/2*x^2 - 4x - 4/(x+1), where are all local and global extrema of f?May be that there are some problems

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the first derivative of f(x) and then you need to find the zeroes of first derivative. The function reaches its extrema at values of x that cancel the derivative.

`f'(x) = -3*4x/(4x^4) - 4 + 4/(x+1)^2`

`` `f'(x) = -3/x^3 - 4 + 4/(x+1)^2 `

`f'(x) = (-3(x+1)^2 - 4x^3(x+1)^2 + 4x^3)/(x^3(x+1)^2)`

`f'(x) = 0 =gt (-3(x+1)^2 - 4x^3(x+1)^2 + 4x^3)/(x^3(x+1)^2) = 0 =gt (-3(x+1)^2 - 4x^3(x+1)^2 + 4x^3) = 0`

 

`-3(x+1)^2 - 4x^3 ((x+1)^2 - 1) = 0`

`-3(x+1)^2 - 4x^3 (x+1-1)(x+1+1) = 0` `-3(x+1)^2 - 4x^3 (x)(x+2) = 0`

 

Notice that sketching the graph of derivative yields a root between (-3,-2), hence the function f(x) has the global extrema at a value of x in (-3,-2).

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