Answer
$x=-\frac{7}{2}, -\frac{1}{2}, 2$
Work Step by Step
Using the rules of derivatives, we find:
$f'(x) = [(2x+7)^6][(x-2)^5]' + [(2x+7)^6]'[(x-2)^5] =(2x+7)^6 * 5 * (x-2)^4 + 6 * 2 * (2x+7)^5(x-2)^5 = (2x+7)^5(x-2)^4[12(x-2) + 5(2x+7)] = (2x+7)^5(x-2)^4(22x+11)$
$y=f(x)$ has horizontal tangents when $f'(x)=0$. Thus, we solve for $x$ from $f'(x) = 0$:
$ (2x+7)^5(x-2)^4(22x+11) = 0$
$x=-\frac{7}{2}, -\frac{1}{2},2$
