/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Locate the absolute extrema of t... [FREE SOLUTION] | 91Ó°ÊÓ

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Locate the absolute extrema of the function on the closed interval. $$f(x)=x^{3}-12 x,[0,4]$$

Short Answer

Expert verified
The absolute maximum of the function \( f(x) = x^{3} - 12x \) on the interval [0,4] is at \( x = 4 \) with value 16, while the absolute minimum is at \( x = 2 \) with value -16.

Step by step solution

01

Derive the function

Start by finding the derivative of the function \(f(x) = x^{3} - 12x\). Apply the power rule, which states that the derivative of \(x^n\) is \(n*x^{n-1}\), to both terms. This gives the derivative as \(f'(x) = 3x^2 - 12\).
02

Check critical points

Next, identify the critical points by setting the derivative equal to zero and solve for \(x\). So, \(3x^2 - 12 = 0\). Solving this equation gives \(x = -2\) and \(x = 2\). However, -2 is not within the given interval [0,4], so we only consider \(x = 2\). Additionally, note the endpoints of the interval, 0 and 4.
03

Evaluate the function

After obtaining the critical points and the endpoints, evaluate the function at these points. So, \(f(0) = 0^3 - 12*0 = 0\), \(f(2) = 2^3 - 12*2 = -16\) and \(f(4) = 4^3 - 12*4 = 16\).
04

Determine the absolute extrema

Now, obtain the absolute extrema by comparing these values. The highest value is called the absolute maximum and the lowest value is the absolute minimum. Here, the absolute maximum of the function on the interval [0,4] is \(f(4) = 16\) and the absolute minimum is \(f(2) = -16\).

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