What is the extreme f(x)=x^2-8ln(x),1<= x<= 5 ?

The extreme f(x)=x^2-8ln(x),1<= x<= 5 is Maximum(1,1),Minimum(2,4-8ln(2)),Maximum(5,25-8ln(5))

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extreme f(x)=x^2-8ln(x),1<= x<= 5

extreme points

f (x) = x^{2} - 8 ln (x), 1 \leq x \leq 5

Maximum (1, 1), Minimum (2, 4 - 8 ln (2)), Maximum (5, 25 - 8 ln (5))

Solution steps

Find the critical points:

x = 1, x = 2, x = 5

Domain of

x^{2} - 8 ln (x), 1 \leq x \leq 5 : 1 \leq x \leq 5

Find intervals:

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < 5

Plug the extreme point

x = 1

into

x^{2} - 8 ln (x) \Rightarrow y = 1

Maximum (1, 1)

Plug the extreme point

x = 2

into

x^{2} - 8 ln (x) \Rightarrow y = 4 - 8 ln (2)

Minimum (2, 4 - 8 ln (2))

Plug the extreme point

x = 5

into

x^{2} - 8 ln (x) \Rightarrow y = 25 - 8 ln (5)

Maximum (5, 25 - 8 ln (5))

Maximum (1, 1), Minimum (2, 4 - 8 ln (2)), Maximum (5, 25 - 8 ln (5))

\frac{x + 5}{x ^{2} - 2 x - 35}

f (x) = x^{2} 5 - x

f (x, y, z) = x^{2} + y^{2}

f (x) = x^{11} - 3 x^{10} + 2

f (x, y) = y^{2} ln (x y)

What is the extreme f(x)=x^2-8ln(x),1<= x<= 5 ?
The extreme f(x)=x^2-8ln(x),1<= x<= 5 is Maximum(1,1),Minimum(2,4-8ln(2)),Maximum(5,25-8ln(5))