What is the extreme f(x)=-x^3-7x^2-16x-42 ?

The extreme f(x)=-x^3-7x^2-16x-42 is Minimum(-8/3 ,-814/27),Maximum(-2,-30)

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extreme f(x)=-x^3-7x^2-16x-42

extreme points

f (x) = - x^{3} - 7 x^{2} - 16 x - 42

Minimum (- \frac{8}{3}, - \frac{814}{27}), Maximum (- 2, - 30)

Solution steps

f^{'} (x) = - 3 x^{2} - 14 x - 16

Find intervals:

Decreasing

: - \infty < x < - \frac{8}{3},

Increasing

: - \frac{8}{3} < x < - 2,

Decreasing

: - 2 < x < \infty

Plug

x = - \frac{8}{3}

into

- x^{3} - 7 x^{2} - 16 x - 42 : - \frac{814}{27}

Minimum (- \frac{8}{3}, - \frac{814}{27})

Plug

x = - 2

into

- x^{3} - 7 x^{2} - 16 x - 42 : - 30

Maximum (- 2, - 30)

Minimum (- \frac{8}{3}, - \frac{814}{27}), Maximum (- 2, - 30)

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

x + \frac{64}{x}

f (x) = x^{3} - 6 x^{2} - 15 x + 10

f (x) = x^{4} - 32 x^{2} + 6

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

What is the extreme f(x)=-x^3-7x^2-16x-42 ?
The extreme f(x)=-x^3-7x^2-16x-42 is Minimum(-8/3 ,-814/27),Maximum(-2,-30)