I am to locate all local and absolute max and mins on the interval $[1,4]$ for $m(x) = \cos(x^2 - 3x)$.
I took the first derivative and got: $-\sin(x^2-3x)(2x-3)$. I then set it to zero to get the critical values, so I set $-\sin(x^2-3x) = 0$ and $2x-3 = 0$. This gave me the following roots: $x = 0, 3$, and $\frac32$.
What confuses me is my textbook does not include $x=0$ as part of the answer and also gives $x = \frac{3 + \sqrt{9+4\pi}}{2}$ as an absolute min. Where did this number come from?
The function $x^2-3x$ takes all the values in $\left[-\frac{9}{4},4\right]$ for $x$ ranging from $1$ to $4$.
In the interval $\left[-\frac{9}{4},4\right]$ there is an absolute maximum of the cosine function at $x=0$ and an absolute minimum at $x=\pi$. It follows that the given function attains its maximum value ($1$) over $[1,4]$ at the points for which $x^2-3x=0$, namely $x=3$, and attains its minimum value ($-1$) at the points for which $x^2-3x=\pi$, namely $x=\frac{3+\sqrt{9+4\pi}}{2}$.
At $x=\frac{3}{2}$ we have a local minimum.