What is the extreme f(x)=8x^3-7ln(x^3),(0.12,7) ?

The extreme f(x)=8x^3-7ln(x^3),(0.12,7) is Minimum((\sqrt[3]{7})/2 ,7-21(1/3 ln(7)-ln(2)))

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extreme f(x)=8x^3-7ln(x^3),(0.12,7)

extreme points

f (x) = 8 x^{3} - 7 ln (x^{3}), (0.12, 7)

Solution steps

Find the critical points:

Domain of

8 x^{3} - 7 ln (x^{3}), (0.12, 7) : 0.12 < x < 7

Find intervals:

Decreasing

Increasing

Plug the extreme point

into

8 x^{3} - 7 ln (x^{3}) \Rightarrow y = 7 - 21 (\frac{1}{3} ln (7) - ln (2))

3 x^{4} - 2

f (x) = \frac{27}{x ^{2} - 4 x - 5}

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x = \frac{4}{3}

y = x^{5} - x^{4}

f (x, y) = 49 - (x - 9)^{2} - y^{2}

What is the extreme f(x)=8x^3-7ln(x^3),(0.12,7) ?
The extreme f(x)=8x^3-7ln(x^3),(0.12,7) is Minimum((\sqrt[3]{7})/2 ,7-21(1/3 ln(7)-ln(2)))