Question

# Consider the function f(x)=2x^{3}-6x^{2}-18x+9 on the interval [-2,4]. What is the absolute minimum of f(x) on [-2,4]? What is the absolute maximum of f(x) on [-2,4]?

Functions
Consider the function $$f(x)=2x^{3}-6x^{2}-18x+9$$ on the interval [-2,4].
What is the absolute minimum of f(x) on [-2,4]?
What is the absolute maximum of f(x) on [-2,4]?

2021-05-20

Step 1
Consider the function
$$f(x)=2x^{3}-6x^{2}-18x+9$$, on [-2,4]
Step 2
The given function is a polynomial so it is continuous everywhere therefore is continuous on [-2,4]
Critical points: Critical points of the function are the points where the derivative of the function either zero or does not exist.
Compute the derivative of the function
$$f(x)=2x^{3}-6x^{2}-18x+9$$
$$f'(x)=6x^{2}-12x-18$$
To find the critical points put $$f'(x)=0.$$
$$f'(x)=0$$
$$6x^{2}-12x-18=0$$
$$6(x^{2}-2x-3)=0$$
$$x^{2}-2x-3=0$$
$$(x-3)(x+1)=0$$
$$x=-1,3$$
There are two critical points $$x=-1$$ and $$x=3$$ and both lies in the interval [-2,4].
Step 3
Evaluate the function value at the critical points and end points.
$$f(x)=2x^{3}-6x^{2}-18x+9$$
at $$x=-2$$
$$f(-2)=2(2)^{3}-6(-2)^{2}-18(-2)+9$$
$$f(-2)=5$$
at $$x=-1$$
$$f(-1)=2(-1)^{3}-6(-1)^{2}-18(-1)+9$$
$$f(-1)=19$$
at $$x=3$$
$$f(3)=2(3)^{3}-6(3)^{2}-18(3)+9$$
$$f(3)=-45$$
at $$x=4$$
$$f(4)=2(4)^{3}-6(4)^{2}-18(4)+9$$
$$f(4)=-31$$
Absolute maximum value and minimum value is the largest and the smallest function value respectively.
Absolute Maximum value $$=19$$ at $$x=-1$$
Absolute Minimum value $$=-45$$ at $$x=-3$$
Step 4
Absolute Maximum value $$=19$$
Absolute Minimum value $$=-45$$