What is the extreme f(x)=4x^5+25x^4-40x^3-1 ?

The extreme f(x)=4x^5+25x^4-40x^3-1 is Maximum(-6,9935),Saddle(0,-1),Minimum(1,-12)

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extreme f(x)=4x^5+25x^4-40x^3-1

extreme points

f (x) = 4 x^{5} + 25 x^{4} - 40 x^{3} - 1

Maximum (- 6, 9935), Saddle (0, - 1), Minimum (1, - 12)

Solution steps

f^{'} (x) = 20 x^{4} + 100 x^{3} - 120 x^{2}

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 6

into

4 x^{5} + 25 x^{4} - 40 x^{3} - 1 : 9935

Maximum (- 6, 9935)

Plug

x = 0

into

4 x^{5} + 25 x^{4} - 40 x^{3} - 1 : - 1

Saddle (0, - 1)

Plug

x = 1

into

4 x^{5} + 25 x^{4} - 40 x^{3} - 1 : - 12

Minimum (1, - 12)

Maximum (- 6, 9935), Saddle (0, - 1), Minimum (1, - 12)

f (x, y) = 6 x y

f (x) = x x^{2} + 25, - 6 \leq x \leq 4

(- 2.3) (1)

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

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

What is the extreme f(x)=4x^5+25x^4-40x^3-1 ?
The extreme f(x)=4x^5+25x^4-40x^3-1 is Maximum(-6,9935),Saddle(0,-1),Minimum(1,-12)