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]?

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asked 2021-05-19
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]?

Answers (1)

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
Answer:
Absolute Maximum value \(=19\)
Absolute Minimum value \(=-45\)

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