Let
$$f(x)=\left\{% \begin{array}{ll} x^3-x^2+10x-5,& x\leq 1 \\ -2x+\log_2 (b^2-2),& x>1 \\ \end{array}% \right.$$
Find all possible real values of $b$ such that $f(x)$ has the greatest value at $x=1$.
How do I proceed? I tried checking LHL=RHL, but I'm getting $b=\pm \sqrt{130}$. The answer is given as $b \in [-\sqrt{130},-2] \cup[\sqrt{2},\sqrt{130}]$.
$$(x^3-x^2+10x-5)'=3x^2-2x+10>0$$ and it's obvious that $f$ decreases on $(1,+\infty).$
Thus, we need $$-2\cdot1+\log_2(b^2-2)\leq f(1)$$ or $$\log_2(b^2-2)\leq7$$ or $$0<b^2-2\leq128$$ or $$[-\sqrt{130},-\sqrt2)\cup(\sqrt2,\sqrt{130}].$$