What is the extreme f(x)=x^4-5x^3 ?

The extreme f(x)=x^4-5x^3 is Saddle(0,0),Minimum(15/4 ,-16875/256)

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extreme f(x)=x^4-5x^3

extreme points

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

Saddle (0, 0), Minimum (\frac{15}{4}, - \frac{16875}{256})

Solution steps

f^{'} (x) = 4 x^{3} - 15 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{15}{4},

Increasing

: \frac{15}{4} < x < \infty

Plug

x = 0

into

x^{4} - 5 x^{3} : 0

Saddle (0, 0)

Plug

x = \frac{15}{4}

into

x^{4} - 5 x^{3} : - \frac{16875}{256}

Minimum (\frac{15}{4}, - \frac{16875}{256})

Saddle (0, 0), Minimum (\frac{15}{4}, - \frac{16875}{256})

f (x) = 1 + \frac{5}{x} - \frac{4}{x ^{2}}

f (x) = \frac{2}{3} x - 5

f (x, y) = x^{3} - x y + 3 y^{2}

f (x) = 4 x^{4} + y^{2} - 2 x y + 1

f (x) = x^{3} - 3 x^{2} - 24 x + 5

What is the extreme f(x)=x^4-5x^3 ?
The extreme f(x)=x^4-5x^3 is Saddle(0,0),Minimum(15/4 ,-16875/256)