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

Answer:

Absolute Maximum value \(=19\)

Absolute Minimum value \(=-45\)