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Chapter 1: Problem 40

Determine the intervals over which the function is increasing, decreasing, or constant. $$ f(x)=x^{2}-4 x $$

Short Answer

Expert verified
The function \(f(x)=x^{2}-4x\) is decreasing for \(x < 2\) and increasing for \(x > 2\). There is no interval where the function is constant.

Step by step solution

01

Find The Derivative of the Function

The first step is to differentiate the function \(f(x)=x^{2}-4x\). Using the power rule, the derivative \(f'(x)\) becomes \(2x-4\).
02

Find The Critical Points

Critical points occur where the derivative of the function is zero or undefined. In this case, to find where \(f'(x) = 0\), set \(2x-4 = 0\). Solving this gives \(x = 2\). This function is defined for all real numbers, so there are no points where the derivative is undefined.
03

Test The Intervals

Now, test the intervals by choosing a number from each of the intervals \(-\infty < x < 2\) and \(2 < x < \infty\), and substituting these numbers into the derivative. Take \(x=0\) for the first interval and \(x=3\) for the second interval. Substituting these into the derivative gives \(f'(0) = -4\) and \(f'(3) = 2\). Therefore, the function is decreasing for \(x < 2\) and increasing for \(x > 2\).

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