Let f(x) = \ cases x ^ 3 - x ^ 2 + 10x - 7, & x ≤ 1 - 2x + _ 2 ( b...

Given, $f$\left(x\right) = \left\{$\begin{cases}{x}^{3} - {x}^{2} + 10x - 7, & x \le 1 \\ - 2x + \log_{2}\left({b}^{2} - 4\right), & x > 1\end{cases}$\right)

Now $f\left(1\right) = 3$

For $x 0$

$⇒ f\left(x\right)$ is increasing

For $x > 1, {f}^{'}\left(x\right) < 0$

$⇒$ function is decreasing.

So, at $x = 1$ there is a point of maxima

So, $\lim_{x \to {1}^{+}}f\left(x\right) = - 2 + \log_{2}\left({b}^{2} - 4\right)$

For maximum value at $x = 1$

$⇒ 3 \ge - 2 + \log_{2}\left({b}^{2} - 4\right)$

$⇒ 32 \ge {b}^{2} - 4 > 0$

$⇒ b \in [ - 6, - 2 ) \cup ( 2, 6 ]$