Question

Easy fun calc problem for medal? :D

Answer 1

Find the exact global maximum & minimum values of the function.

\[f(x)=x-\ln(x) \]
for x>0

Answer 2

\[\large f(x) = x - \ln(x)\]
\[\large f \ ' (x) = 1 - \frac{1}{x}\]
\[\large f \ ' (x) = \frac{x}{x}-\frac{1}{x}\]
\[\large f \ ' (x) = \frac{x-1}{x}\]

Set f ' (x) equal to zero and solve for x to find any critical values. Keep in mind that x > 0

\[\large f \ ' (x) = 0\]
\[\large \frac{x-1}{x} = 0\]
\[\large x-1 = 0\]
\[\large x = 1\]

Since x = 1 satisfies x > 0, this means we have a valid critical value in the proper domain. Let's see if we have a sign change

If we pick a value smaller than x = 1, say x = 0.5, then

\[\large f \ ' (x) = \frac{x-1}{x}\]
\[\large f \ ' (0.5) = \frac{0.5-1}{0.5}\]
\[\large f \ ' (0.5) = \frac{-0.5}{0.5}\]
\[\large f \ ' (0.5) = -1\]

which means f ' (x) is negative and f(x) is decreasing on the interval 0 < x < 1

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Repeat the last few steps but now use a value larger than 1, say 1.5

\[\large f \ ' (x) = \frac{x-1}{x}\]
\[\large f \ ' (1.5) = \frac{1.5-1}{0.5}\]
\[\large f \ ' (1.5) = \frac{0.5}{1.5}\]
\[\large f \ ' (1.5) \approx 0.333\]

which indicates f(x) is increasing on the interval x > 1

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As we pass through x = 1, f(x) transitions from decreasing to increasing like so



which suggests this is a local min. It's possible this is also a global min as well.

There is no local max since we only have one critical value. This one critical value corresponds to a local min.

Recall that for global extrema, we need to test endpoints. This interval is 0 < x < infinity which means there aren't really any endpoints to test. So we can ignore this part.

The only thing to pull away from all this is that x = 1 is the critical value that leads to a local min. In this case, it's also a global min because it yields the smallest output of any input in the domain. This graph in the attachment confirms the answer.

Answer 3

Almost forgot: To find the global min output, you plug in x = 1 back into the f(x) function

f(x) = x - ln(x)
f(1) = 1 - ln(1)
f(1) = 1-0
f(1) = 1

so in this case, the input and output are the same coincidentally