What is the extreme f(x)=x^{7/5}-12x^{1/5} ?

The extreme f(x)=x^{7/5}-12x^{1/5} is Minimum((12/7)^{5/6},(12/7)^{7/6}-12(12/7)^{1/6})

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extreme f(x)=x^{7/5}-12x^{1/5}

extreme points

f (x) = x^{\frac{7}{5}} - 12 x^{\frac{1}{5}}

Minimum ((\frac{12}{7})^{\frac{5}{6}}, (\frac{12}{7})^{\frac{7}{6}} - 12 (\frac{12}{7})^{\frac{1}{6}})

Solution steps

f^{'} (x) = \frac{7 x ^{\frac{6}{5}} - 12}{5 x ^{\frac{4}{5}}}

Find intervals:

Increasing

: - \infty < x < - (\frac{12}{7})^{\frac{5}{6}},

Decreasing

: - (\frac{12}{7})^{\frac{5}{6}} < x < 0,

Decreasing

: 0 < x < (\frac{12}{7})^{\frac{5}{6}},

Increasing

: (\frac{12}{7})^{\frac{5}{6}} < x < \infty

Plug

x = (\frac{12}{7})^{\frac{5}{6}}

into

x^{\frac{7}{5}} - 12 x^{\frac{1}{5}} : (\frac{12}{7})^{\frac{7}{6}} - 12 (\frac{12}{7})^{\frac{1}{6}}

Minimum ((\frac{12}{7})^{\frac{5}{6}}, (\frac{12}{7})^{\frac{7}{6}} - 12 (\frac{12}{7})^{\frac{1}{6}})

f (x) = x^{2} - 12

f (x) = 4 x^{\frac{3}{5}} - x^{\frac{4}{5}}

f (x) = \frac{x + 3}{x ^{2} - x - 12}

f (x) = - 5 sin (x) cos (x)

x^{4} + 8 x^{3}

What is the extreme f(x)=x^{7/5}-12x^{1/5} ?
The extreme f(x)=x^{7/5}-12x^{1/5} is Minimum((12/7)^{5/6},(12/7)^{7/6}-12(12/7)^{1/6})