I am given a function $$f(x) = \begin{cases} x^3 - x^2 + 10x - 5, & \text{if x $\leqslant$ 1} \\[2ex] -2x + log_2(b^2 - 2), & \text{if x $\geqslant$ 1} \end{cases}$$ I need to find the range of values of b so that f(x) has its maximum value at x = 1.
For this, I observe that the first part of the function is always increasing. Since it is not mentioned that the given function is continuous, and moreover the second function is continuously decreasing, hence I need to ensure that the value of the second function at x = 1 is less than the value of the first function at x = 1.
Am I correct? Or is there any flaw in my argument since I am getting the wrong answer according to my book.
You need to see the behavior of the function when $x<1$ and when $x>1$. Observe the following case:
Note that for $x<1$, we have $f'(x) = 3x^{2} - 2x + 10$ which is strictly increasing and thus for any $x<1$, we know that $f(x)<f(1)$.
Now, for $x>1$, we know that $f'(x) = -2$ and therefore $f$ is strictly decreasing for any $x>1$.
Hence, the only possible solution for $f$ to have a maximum value is no other than $x=1$. However, from your definition, we know that $f(1)= 5 = -2+ \log_{2}(b^{2}-2)$. Therefore, you can get the result (by straightforward calculation) that either $b=\sqrt{130}$ or $b=-\sqrt{130}$.
I leave the detail of why $f'(x)$ is strictly increasing when $x<1$ but is strictly decreasing when $x>1$ as an exercise for you. The rest of the calculation is also easy so I am sure you can do it.