What is the extreme f(x)=3x^5-20x^3+42 ?

The extreme f(x)=3x^5-20x^3+42 is Maximum(-2,106),Saddle(0,42),Minimum(2,-22)

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extreme f(x)=3x^5-20x^3+42

extreme points

f (x) = 3 x^{5} - 20 x^{3} + 42

Maximum (- 2, 106), Saddle (0, 42), Minimum (2, - 22)

Solution steps

f^{'} (x) = 15 x^{4} - 60 x^{2}

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

3 x^{5} - 20 x^{3} + 42 : 106

Maximum (- 2, 106)

Plug

x = 0

into

3 x^{5} - 20 x^{3} + 42 : 42

Saddle (0, 42)

Plug

x = 2

into

3 x^{5} - 20 x^{3} + 42 : - 22

Minimum (2, - 22)

Maximum (- 2, 106), Saddle (0, 42), Minimum (2, - 22)

f (x, y) = 3 x^{2} - y^{2} - 7 x - y + 8

f (x) = 2 + 28 x - 2 x^{2}

f (x) = 2 x^{3} - x^{2} - 4 x + 10, - 1 \leq x \leq 0

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

2 sin (\frac{θ}{2})

What is the extreme f(x)=3x^5-20x^3+42 ?
The extreme f(x)=3x^5-20x^3+42 is Maximum(-2,106),Saddle(0,42),Minimum(2,-22)