How do you find the critical numbers for $f (x) = x - 2 ln (x)$ $f(x) = x-2ln(x)$ to determine the maximum and minimum?

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How do you find the critical numbers for $f (x) = x - 2 ln (x)$ $f(x) = x-2ln(x)$ to determine the maximum and minimum?

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Answer 1 · 2017-08-13T14:52:03

The first derivative is given by

$f'(x) = 1 - 2/x$

Which has critical points at $0$ and when $f'(x) = 0$ .

$0 = 1- 2/x -> 2/x = 1 -> x = 2$

However, $x = 0$ is not really a critical point because the initial function is undefined there. Recall that $ln(0) = O/$ . Now let's see if $x = 2$ is a maximum or a minimum. At $x = 1$ , the function is decreasing because $f'(1) < 0$ . Hence, $x = 2$ will be an absolute minimum.

Hopefully this helps!

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How do you find the critical numbers for $f (x) = x - 2 ln (x)$ $f(x) = x-2ln(x)$ to determine the maximum and minimum?

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How do you find the critical numbers for f(x) = x-2ln(x)f(x)=x−2ln(x)f(x) = x-2ln(x) to determine the maximum and minimum?

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How do you find the critical numbers for $f (x) = x - 2 ln (x)$ $f(x) = x-2ln(x)$ to determine the maximum and minimum?