Let $f(x)= \begin{cases} x^3-x^2+10 x-7, & x \leq 1 \\ -2 x+\log _2\left(b^2-4\right), & x>1\end{cases}$ Then the set of all values of $b$, for which $f(x)$ has maximum value at $x=1$, is :
To determine the set of all values of \(b\), for which the function \(f(x)\) has its maximum value at \(x=1\), we need to analyze the behavior of \(f(x)\) at that point and the surrounding interval.
For \(x > 1\), to satisfy the condition that \(f(x)\) has a maximum value at \(x=1\), \(-2x + \log_2(b^2 - 4)\) for \(x > 1\) must be less than or equal to 3 when \(x=1\).
Calculate the derivative of both parts to check behavior around \(x = 1\):
For \(x \leq 1\), the derivative \(f'(x) = 3x^2 - 2x + 10\). At \(x = 1\), \(f'(1) = 3(1)^2 - 2(1) + 10 = 11\), which indicates that the left side has a positive slope, meaning it increases to \(x=1\).
For \(x > 1\), the derivative \(f'(x) = -2\), constant and negative, hence decreasing for \(x > 1\).
To ensure a maximum at \(x=1\), the function must have its peak at the transition, hence we must adjust:
Set \(-2x + \log_2(b^2-4) \leq 3\).
At \(x = 1\), \(-2(1) + \log_2(b^2-4) = 3\) simplifies to \(\log_2(b^2-4) = 5\).
Solve for \(b^2 - 4 = 2^5 = 32\) so \(b^2 = 36\).
This gives possible \(b = \pm 6\).
When \(b = 6\), \(-2(1) + \log_2(32) = 3\), maintaining maximum at \(x=1\).
When \(b = -6\), same logic applies.
Therefore, based on \(\log_2(b^2 - 4)\), \(-\sqrt{36}=-6\) to \(\sqrt{36}=6\).
Ensure \(b^2 > 4\), so exclude ends of intervals where equation is \((b^2=4)\).
The correct set where maximum occurs at \(x = 1\) is \([-6,-2) \cup (2,6]\).
