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Chapter 5: Problem 69

Find any relative extrema of the function. \(f(x)=\operatorname{arcsec} x-x\)

Short Answer

Expert verified
The function \(f(x)=\operatorname{arcsec} x-x\) has a relative maximum at \(x=-1\) and a relative minimum at \(x=1\)

Step by step solution

01

Find the derivative

The derivative of the function \(f(x)=\operatorname{arcsec} x-x\) is given by using the chain rule and knowing that the derivative of \(x\) is \(1\) and the derivative of \(\operatorname{arcsec} x\) is \(\frac {1} {|x|\sqrt {x^{2}-1}}\), for \(|x|>1\). So, \(f'(x)=\frac {1}{|x|\sqrt {x^{2}-1}}-1\)
02

Find the critical points

The critical points of a function are the points at which the derivative of the function is either 0 or undefined. So, we set \(f'(x)=0\) and \(f'(x)\) undefined to solve for \(x\). Here \(f'(x)\) is undefined at \(x=1\) and \(x=-1\) and there is no real solution for \(f'(x)=0\). So we have critical points at \(x=-1\) and \(x=1\)
03

Apply the first derivative test

So now we have three intervals to consider, they are, \((-鈭�,-1)\), \((-1,1)\), and \((1, 鈭�)\). We pick a test point within each interval and evaluate the sign of \(f'(x)\) at those points. We'll see that \(f'(x)>0\) for \(x<-1\), \(f'(x)<0\) for \(-10\) for \(1<x\)
04

Verify relative extrema

As the derivative changes from positive to negative at \(x=-1\), this is a point of relative maximum. And, as the derivative changes from negative to positive at \(x=1\), this is a point of relative minimum.

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