Question

2.Find the maximum and minimum points for the function f (x) =x³ - 3x² - 9x - 8 (Differentiation)

Answer 1

Solution. Find the derivative of the function f(x) using the rules of differentiation

$$f'(x)=3x^2-6x-9.$$

Equate the derivative to zero and find the roots of the equation

$$3x^2-6x-9=0$$ $$x^2-2x-3=0.$$

$$D=(-2)^-4\times1\times(-3)=4+12=16$$ $$x_1=\frac{2-\sqrt{16}}{2}=\frac{2-4}{2}=-1$$

$$x_2=\frac{2+\sqrt{16}}{2}=\frac{2+4}{2}=3$$

Points x = -1 and x = 3 divide the domain into three intervals. Let us find the sign of the derivative on each of the intervals.

For

$$x\in(-\infin;-1)$$ $$f'(-2)=3\times(-2)^2-6\times(-2)-9=12+12-9=15>0.$$

Since the derivative f'(x)>o the function f(x) increases on the specified interval.

For

$$x\in(-1;3)$$

$$f'(-2)=3\times0^2-6\times0-9=0-0-9=-9<0.$$

Since the derivative f'(x)<o the function f(x) decreases on the specified interval.

For

$$x\in(3;+\infin)$$

$$f'(4)=3\times4^2-6\times4-9=48-24-9=15>0.$$

Since the derivative f'(x)>o the function f(x) increases on the specified interval.

As result get at the point x=-1

$$f(-1)=(-1)^3-3\times(-1)^2-9\times(-1)-8=-1-3+9-8=-3$$

the maximum of the function f(x);

at the point x=3

$$f(3)=3^3-3\times3^2-9\times3-8=27-27+27-8=-35$$

the minimum of the function f(x).

Answer. Point x=-1 f(-1)=-3 is the maximum of the function f(x); point x=3 f(-1)=-35 is the minimum of the function f(x).