(a) Find the domain of the funtion $f\left(x\right)={x}^{3}-3{x}^{2}-9x,\text{}x\in R$ Finde the derivative ${f}^{\prime}\left(x\right)=3{x}^{2}-6x-9$ Find the domain of derivative ${f}^{\prime}\left(x\right)=3{x}^{2}-6x-9,\text{}x\in R$ Substitute f'(x)=0 $0=3{x}^{2}-6x-9$ Solve equation $x=-1$

$x=3$

Determine th intervals $\u27e8-\mathrm{\infty},-1\u27e9,\u27e8-1,3\u27e9$

$\u27e8-1,3\u27e9,\u27e83,+\mathrm{\infty}\u27e9$ Choose the points ${x}_{1}=-2$

${x}_{2}=0$

${x}_{3}=0$

${x}_{4}=0$ Calculate the values of the derivatives ${f}^{\prime}(-2)=15$

${f}^{\prime}\left(0\right)=-9$

${f}^{\prime}\left(0\right)=-9$

${f}^{\prime}\left(4\right)=15$ The relative maximum at x=-1 The relative minimum is at x=3 ${f}^{\prime}\left(x\right)=3{x}^{2}-6x-9,\text{}x=-1$

${f}^{\prime}\left(x\right)=3{x}^{2}-6x-9,\text{}x=3$ Calculate the value $f(-1)=5$

f(3)=-27 The relative maximum of the function is 5 at x=-1 The relative minimum of the function is -27 at x=3 (b) To find x-intercept, substitute f(x)=0 $0={x}^{3}-3{x}^{2}-9x$ Swap the sides ${x}^{3}-3{x}^{2}-9x=0$ Factor the expression $x\times ({x}^{2}-3x-9)=0$ Separate onto possible cases $x=0$

${x}^{2}-3x-9=0$ Solve the equation $x=0$

$x=\frac{3+3\sqrt{5}}{2}$

$x=\frac{3-3\sqrt{5}}{2}$ There is 3 solutions (c) and (d) Let's plot the polynomial to determine the intervals of increasing and decreasing. Increase by: $(-\mathrm{\infty},-1),(3,\mathrm{\infty})$ Decreases by: $(-1,3)$