What is the extreme y=3x^4+16x^3=x^3(3x+16) ?

The extreme y=3x^4+16x^3=x^3(3x+16) is Minimum(-4,-256),Saddle(0,0)

extreme y=3x^4+16x^3=x^3(3x+16)

extreme points

y = 3 x^{4} + 16 x^{3} = x^{3} (3 x + 16)

Minimum (- 4, - 256), Saddle (0, 0)

Solution steps

f^{'} (x) = 12 x^{2} (x + 4)

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 4

into

x^{3} (3 x + 16) : - 256

Minimum (- 4, - 256)

Plug

x = 0

into

x^{3} (3 x + 16) : 0

Saddle (0, 0)

Minimum (- 4, - 256), Saddle (0, 0)

f (x) = - 4 x^{3} - 6 x^{2} + 72 x + 6

y = 12 x^{\frac{5}{7}} + 5 x^{\frac{12}{7}}

y = x^{3} - 3 x^{2} + 7 x - 10

f (x) = - \frac{10}{3} x^{3} - \frac{31}{2} x^{2} - 3 x + 4

f (x) = \frac{x ^{2} + 7 x + 13}{x + 4}

What is the extreme y=3x^4+16x^3=x^3(3x+16) ?
The extreme y=3x^4+16x^3=x^3(3x+16) is Minimum(-4,-256),Saddle(0,0)