What is the extreme f(x)= 1/x+ln(x),0.5<= x<= 4 ?

The extreme f(x)= 1/x+ln(x),0.5<= x<= 4 is Maximum(0.5,1.30685…),Minimum(1,1),Maximum(4, 1/4+2ln(2))

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extreme f(x)= 1/x+ln(x),0.5<= x<= 4

extreme points

f (x) = \frac{1}{x} + ln (x), 0.5 \leq x \leq 4

Maximum (0.5, 1.30685 \dots), Minimum (1, 1), Maximum (4, \frac{1}{4} + 2 ln (2))

Solution steps

Find the critical points:

x = 0.5, x = 1, x = 4

Domain of

\frac{1}{x} + ln (x), 0.5 \leq x \leq 4 : 0.5 \leq x \leq 4

Find intervals:

Decreasing

: 0.5 < x < 1,

Increasing

: 1 < x < 4

Plug the extreme point

x = 0.5

into

\frac{1}{x} + ln (x) \Rightarrow y = 1.30685 \dots

Maximum (0.5, 1.30685 \dots)

Plug the extreme point

x = 1

into

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

Minimum (1, 1)

Plug the extreme point

x = 4

into

\frac{1}{x} + ln (x) \Rightarrow y = \frac{1}{4} + 2 ln (2)

Maximum (4, \frac{1}{4} + 2 ln (2))

Maximum (0.5, 1.30685 \dots), Minimum (1, 1), Maximum (4, \frac{1}{4} + 2 ln (2))

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y = x + 1 - x

f (x) = x^{2} + 6 x + 6

e^{5 x} + e^{- x}

What is the extreme f(x)= 1/x+ln(x),0.5<= x<= 4 ?
The extreme f(x)= 1/x+ln(x),0.5<= x<= 4 is Maximum(0.5,1.30685…),Minimum(1,1),Maximum(4, 1/4+2ln(2))