What is the extreme f(x)=-7x^3+21x+4 ?

The extreme f(x)=-7x^3+21x+4 is Minimum(-1,-10),Maximum(1,18)

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extreme f(x)=-7x^3+21x+4

extreme points

f (x) = - 7 x^{3} + 21 x + 4

Minimum (- 1, - 10), Maximum (1, 18)

Solution steps

f^{'} (x) = - 21 x^{2} + 21

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

- 7 x^{3} + 21 x + 4 : - 10

Minimum (- 1, - 10)

Plug

x = 1

into

- 7 x^{3} + 21 x + 4 : 18

Maximum (1, 18)

Minimum (- 1, - 10), Maximum (1, 18)

x^{3} - 2 x^{2} - 4 x + 4

f (x) = 16 x + 17 x^{\frac{16}{17}}

f (x) = x^{\frac{5}{7}}

y = 9 sin (∣ x ∣), - 2 π \leq x \leq 2 π

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

What is the extreme f(x)=-7x^3+21x+4 ?
The extreme f(x)=-7x^3+21x+4 is Minimum(-1,-10),Maximum(1,18)