1. **find the location of all critical points for each function below:\na) (f(x)=3x^{2}-9x + 12)\nb)…

Explanation:

Step1: Recall critical - point definition

Critical points occur where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined.

Step2: Differentiate $f(x)=3x^{2}-9x + 12$

Using the power rule $(x^n)^\prime=nx^{n - 1}$, we have $f^{\prime}(x)=\frac{d}{dx}(3x^{2}-9x + 12)=6x-9$.

Step3: Set $f^{\prime}(x)=0$ for part a

$6x - 9=0$. Solving for $x$ gives $6x=9$, so $x=\frac{9}{6}=\frac{3}{2}$.

Step4: Differentiate $f(x)=\sin(2x)$

Using the chain - rule, if $y = \sin(u)$ and $u = 2x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}(\sin(u))=\cos(u)$ and $\frac{d}{dx}(2x)=2$, we have $f^{\prime}(x)=2\cos(2x)$.

Step5: Set $f^{\prime}(x)=0$ for part b

$2\cos(2x)=0$, so $\cos(2x)=0$. Then $2x=\frac{\pi}{2}+k\pi$, where $k\in\mathbb{Z}$. Solving for $x$ gives $x=\frac{\pi}{4}+\frac{k\pi}{2}$. For $-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}$, when $k = - 1$, $x=\frac{\pi}{4}-\frac{\pi}{2}=-\frac{\pi}{4}$; when $k = 0$, $x=\frac{\pi}{4}$; when $k = 1$, $x=\frac{\pi}{4}+\frac{\pi}{2}=\frac{3\pi}{4}\gt\frac{\pi}{2}$ (rejected).

Answer:

a) $x = \frac{3}{2}$ b) $x=-\frac{\pi}{4},\frac{\pi}{4}$