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

1 Answer | Add Yours

Top Answer

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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).

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question