Chapter 4.1, Problem 2QCE

Let $f\left(x\right)=0.1\left(x^3-3x^2-9x\right)$ . Then $\begin{array}{c}{f}^{\prime }\left(x\right)=0.1\left(3x^2-6x-9\right)=0.3\left(x+1\right)\left(x-3\right)\\ f^{″}\left(x\right)=0.6\left(x-1\right)\end{array}$ f

(a) Solution to $f^{\prime }\left(x\right)=0$ are $x=\text{\_\_\_\_\_\_\_}$ .

(b) The function $f$ is increasing on the interval (s) $\text{\_\_\_\_\_\_\_}$ .

(c) The function $f$ is concave down on the interval (s) $\text{\_\_\_\_\_\_\_}$ .

(d) $\text{\_\_\_\_\_\_\_}$ is an inflection point on the graph of $f$ .