Need help with a calculus maximum question

Question

The question is:

Which values of $k$ give a maximum at $x=−1$ for $f(x)=(k+1)x^4−(3k+2)x^2−2kx$?

I found that $f'(x)=4x^3k+4x^3-6xk-4x-2k$, but I'm confused on where to go from here because plugging in $-1$ leaves $f'(x)=0$.

Answer 1

If we consider $$f(x)=(k+1)x^4−(3k+2)x^2−2kx$$ $$f'(x)=4 (k+1) x^3-2 (3 k+2) x-2 k$$ $$f''(x)=12 (k+1) x^2-2 (3 k+2)$$ then $$f(-1)=-1 \qquad f'(-1)=0 \qquad f''(-1)=6k+8$$

So, in order the point be a maximum, you need $6k+8 <0$ (see gammatester's comment).

Answer 2

No systematic solution, but playing with the functions shows that for $k=-2$ you have

$$f(x) = -x^4+4x^2+4x, \quad f(-1)=-1$$ $$f'(x) = -4x^3+8x+4, \quad f'(-1) = 0$$ $$f''(x) = -12x^2+8,\quad f''(-1) = -4$$

so there is local maximum for $k=-2$ at $(-1, -1).$