Question

Find the maximum and minimum values of y = 1/x + lnx on the interval 1/e <= x <= e.

Answer 1

if y = 1/x + ln(x), what is y' ?

Answer 2

y' would be \[\frac{ -1 }{ x^2 } + \frac{ 1 }{ x }\] which equals \[\frac{ x-1 }{ x^2 }\] right?

Answer 3

@jim_thompson5910

Answer 4

both statements are correct

Answer 5

ok, then how would I find the maximum and minimum values?

Answer 6

That's the part of the problem that I don't understand how to do

Answer 7

once you get y', you set it equal to 0 and solve for x

Answer 8

what do you get when you do that

Answer 9

x = 1

Answer 10

so that is a critical value

Answer 11

we now have to determine if there is a local min, local max, or saddle point at x = 1

Answer 12

do you know how to check?

Answer 13

no, my teacher did not teach me any of this, sorry :( Can you please explain to me how to check for those things?

Answer 14

Draw a number line. I'm only going to mark the point x = 1 on this number line

Answer 15

Now let's pick a number to the right of the number line, say x = 2

Answer 16

plug this into y' and evalute

y ' = (x-1)/(x^2)

y ' = (2-1)/(2^2)

y' = 1/4

notice how the result is positive. You can pick ANY value to the right of x = 1, plug it into y' and it will be a positive result. So we put a bit plus sign over the the right side of x = 1



that plus sign says that the derivative is positive on that interval (from 1 to infinity)

Answer 17

now pick a value to the left of x = 1, I'll do x = 0.5

y ' = (x-1)/(x^2)

y ' = (0.5-1)/(0.5^2)

y ' = -2

now that's negative

so y' is negative on the interval from 0 to 1.