For the function f(x)=x(x2+1)2 on [-2,2] Find the critical points and the absolute extreme values of f on the given interval.

Question

a better view of the question

Answer 1

The correct minimum is (-1/√3 or -√3/3,-3√3/16) and the correct maximum is (1/√3 or √3/3,3√3/16). This solution is not listed in the options.

Here’s the full solution.

f(x)=x/(x²+1)², df/dx=[(x²+1)²-(x)(2(x²+1))(2x)]/(x²+1)⁴=

(x⁴+2x²+1-4x²(x²+1))/(x²+1)²=0 at critical points.

-3x⁴-2x²+1=0=-(3x²-1)(x²+1), so x²=⅓ and x=√⅓ or -√⅓, which can be written √3/3, -√3/3.

f(0)=0, so we can see which is a min and which is a max:

f(√⅓)=3√3/16 (max), f(-√⅓)=-3√3/16 (min).

 

Comments

You really do need to tell your tutor as soon as possible about these unacceptable errors in these questions, for your own sake and for the sake of thousands of other students. Yes, some errors are inevitable, but this level of errors is unacceptable. It could affect the future careers of many students who may be confused and wasting time (especially in exam conditions) trying to fit their solutions to one of the options.