What is the extreme f(x)=12x^5+30x^4-160x^3+5 ?

The extreme f(x)=12x^5+30x^4-160x^3+5 is Maximum(-4,5637),Saddle(0,5),Minimum(2,-411)

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extreme f(x)=12x^5+30x^4-160x^3+5

extreme points

f (x) = 12 x^{5} + 30 x^{4} - 160 x^{3} + 5

Maximum (- 4, 5637), Saddle (0, 5), Minimum (2, - 411)

Solution steps

f^{'} (x) = 60 x^{4} + 120 x^{3} - 480 x^{2}

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 4

into

12 x^{5} + 30 x^{4} - 160 x^{3} + 5 : 5637

Maximum (- 4, 5637)

Plug

x = 0

into

12 x^{5} + 30 x^{4} - 160 x^{3} + 5 : 5

Saddle (0, 5)

Plug

x = 2

into

12 x^{5} + 30 x^{4} - 160 x^{3} + 5 : - 411

Minimum (2, - 411)

Maximum (- 4, 5637), Saddle (0, 5), Minimum (2, - 411)

f (x) = 2 x^{3} + 16 x^{2} + 32 x + 5

y = \frac{x + 5}{x ^{2} - 4}

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

f (x) = x^{2} - 16 x, - 4 \leq x \leq 4

f (x) = e^{x} (x + 4)

What is the extreme f(x)=12x^5+30x^4-160x^3+5 ?
The extreme f(x)=12x^5+30x^4-160x^3+5 is Maximum(-4,5637),Saddle(0,5),Minimum(2,-411)