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Chapter 4: Problem 3

Find the critical points and the local extreme values. $$f(x)=x+\frac{1}{x}$$.

Short Answer

Expert verified
The critical points of the function are \(x=1\) and \(x=-1\). The local minimum value is 2 at \(x=1\) and the local maximum value is -2 at \(x=-1\).

Step by step solution

01

Calculating the derivative of the function

The derivative of the function \(f(x)\) is computed using the rules of differentiation, which yields \(f'(x)=1-\frac{1}{x^2}\).
02

Finding the critical points

The critical points are found by setting the derivative equal to zero and solving for \(x\). So, \(1-\frac{1}{x^2}=0 \Rightarrow x^2=1\). The solutions are \(x_1=1\) and \(x_2=-1\), which are the critical points of the function.
03

Calculating the second derivative of the function

By taking the derivative of \(f'(x)\), we get \(f''(x)=\frac{2}{x^3}\).
04

Applying the Second Derivative Test

If \(f''(x) > 0\) at a critical point, then the function has a local minimum at that point. If \(f''(x) < 0\) then the function has a local maximum. Here, \(f''(1) = 2 > 0\) and \(f''(-1) = -2 < 0\), so the function \(f(x)\) has a local minimum at \(x=1\) and a local maximum at \(x=-1\).
05

Identifying the local extreme values

Plug the critical points into the original function to find the local extreme values: \(f(1) = 1+1=2\) and \(f(-1) = -1 - 1 = -2\). So, the local minimum is 2 at \(x=1\) and local maximum is -2 at \(x=-1\).

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