Let f(x) = begin{cases} x^{3}-x^{2}+10x-7, & | vedclass

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(C) First,we evaluate $f(1) = 1^{3} - 1^{2} + 10(1) - 7 = 1 - 1 + 10 - 7 = 3$.For $x < 1$,the derivative is $f'(x) = 3x^{2} - 2x + 10$. The discrimina

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$[-6, -2) \cup (2, 6]$

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(C) First,we evaluate $f(1) = 1^{3} - 1^{2} + 10(1) - 7 = 1 - 1 + 10 - 7 = 3$.For $x  0$,$f'(x) > 0$ for all $x For $x > 1$,$f(x) = -2x + \log_{2}(b^{2}-4)$. For $f(x)$ to have a maximum at $x=1$,the function must be non-increasing for $x > 1$ near $x=1$,and the limit as $x 	o 1^{+}$ must be less than or equal to $f(1)$.Condition $1$: The argument of the logarithm must be positive,so $b^{2} - 4 > 0$,which implies $b \in (-\infty, -2) \cup (2, \infty)$.Condition $2$: The limit as $x 	o 1^{+}$ must satisfy $\lim_{x 	o 1^{+}} f(x) \leq f(1)$.$-2(1) + \log_{2}(b^{2}-4) \leq 3$$\log_{2}(b^{2}-4) \leq 5$$b^{2} - 4 \leq 2^{5} = 32$$b^{2} \leq 36$$|b| \leq 6$,so $b \in [-6, 6]$.Combining the conditions $b \in (-\infty, -2) \cup (2, \infty)$ and $b \in [-6, 6]$,we get $b \in [-6, -2) \cup (2, 6]$.

Let $f(x) = \begin{cases} x^{3}-x^{2}+10x-7, & x \leq 1 \\ -2x+\log_{2}(b^{2}-4), & x > 1 \end{cases}$. Then the set of all values of $b$,for which $f(x)$ has a maximum value at $x=1$,is:

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