What is the extreme f(x)= 1/x+ln(x),e^{-1}<= x<= e ?

The extreme f(x)= 1/x+ln(x),e^{-1}<= x<= e is Maximum(e^{-1},e-1),Minimum(1,1),Maximum(e, 1/e+1)

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extreme f(x)= 1/x+ln(x),e^{-1}<= x<= e

extreme points

f (x) = \frac{1}{x} + ln (x), e^{- 1} \leq x \leq e

Maximum (e^{- 1}, e - 1), Minimum (1, 1), Maximum (e, \frac{1}{e} + 1)

Solution steps

Find the critical points:

x = e^{- 1}, x = 1, x = e

Domain of

\frac{1}{x} + ln (x), e^{- 1} \leq x \leq e : e^{- 1} \leq x \leq e

Find intervals:

Decreasing

: e^{- 1} < x < 1,

Increasing

: 1 < x < e

Plug the extreme point

x = e^{- 1}

into

\frac{1}{x} + ln (x) \Rightarrow y = e - 1

Maximum (e^{- 1}, e - 1)

Plug the extreme point

x = 1

into

\frac{1}{x} + ln (x) \Rightarrow y = 1

Minimum (1, 1)

Plug the extreme point

x = e

into

\frac{1}{x} + ln (x) \Rightarrow y = \frac{1}{e} + 1

Maximum (e, \frac{1}{e} + 1)

Maximum (e^{- 1}, e - 1), Minimum (1, 1), Maximum (e, \frac{1}{e} + 1)

f (x) = x^{2} - y^{2} - x - y

f (x) = x + \frac{1}{x}, (0.2, 4)

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 17

x^{2} - 6 x + 7

f (x) = x^{3} + 6 x^{2} + 9 x - 8

What is the extreme f(x)= 1/x+ln(x),e^{-1}<= x<= e ?
The extreme f(x)= 1/x+ln(x),e^{-1}<= x<= e is Maximum(e^{-1},e-1),Minimum(1,1),Maximum(e, 1/e+1)