In each item, do the following analytically: (a) find the relative extrema of f. (b)...

(a) Find the domain of the funtion ${\displaystyle f\left(x\right)=x^3-3x^2-9x,x\in R}$ Finde the derivative ${\displaystyle f^{\prime }\left(x\right)=3x^2-6x-9}$ Find the domain of derivative ${\displaystyle f^{\prime }\left(x\right)=3x^2-6x-9,x\in R}$ Substitute f'(x)=0 ${\displaystyle 0=3x^2-6x-9}$ Solve equation ${\displaystyle x=-1}$
${\displaystyle x=3}$
Determine th intervals ${\displaystyle ⟨-\mathrm{\infty },-1⟩,⟨-1,3⟩}$
${\displaystyle ⟨-1,3⟩,⟨3,+\mathrm{\infty }⟩}$ Choose the points ${\displaystyle x_1=-2}$
${\displaystyle x_2=0}$
${\displaystyle x_3=0}$
${\displaystyle x_4=0}$ Calculate the values of the derivatives ${\displaystyle f^{\prime }(-2)=15}$
${\displaystyle f^{\prime }\left(0\right)=-9}$
${\displaystyle f^{\prime }\left(0\right)=-9}$
${\displaystyle f^{\prime }\left(4\right)=15}$ The relative maximum at x=-1 The relative minimum is at x=3 ${\displaystyle f^{\prime }\left(x\right)=3x^2-6x-9,x=-1}$
${\displaystyle f^{\prime }\left(x\right)=3x^2-6x-9,x=3}$ Calculate the value ${\displaystyle f(-1)=5}$
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 ${\displaystyle 0=x^3-3x^2-9x}$ Swap the sides ${\displaystyle x^3-3x^2-9x=0}$ Factor the expression ${\displaystyle x\times (x^2-3x-9)=0}$ Separate onto possible cases ${\displaystyle x=0}$
${\displaystyle x^2-3x-9=0}$ Solve the equation ${\displaystyle x=0}$
${\displaystyle x=\frac{3+3\sqrt{5}}{2}}$
${\displaystyle x=\frac{3-3\sqrt{5}}{2}}$ There is 3 solutions (c) and (d) Let's plot the polynomial to determine the intervals of increasing and decreasing. Increase by: ${\displaystyle (-\mathrm{\infty },-1),(3,\mathrm{\infty })}$ Decreases by: ${\displaystyle (-1,3)}$