What is the extreme f(x)=-2x^3+42x^2-240x+5 ?

The extreme f(x)=-2x^3+42x^2-240x+5 is Minimum(4,-411),Maximum(10,-195)

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extreme f(x)=-2x^3+42x^2-240x+5

extreme points

f (x) = - 2 x^{3} + 42 x^{2} - 240 x + 5

Minimum (4, - 411), Maximum (10, - 195)

Solution steps

f^{'} (x) = - 6 x^{2} + 84 x - 240

Find intervals:

Decreasing

: - \infty < x < 4,

Increasing

: 4 < x < 10,

Decreasing

: 10 < x < \infty

Plug

x = 4

into

- 2 x^{3} + 42 x^{2} - 240 x + 5 : - 411

Minimum (4, - 411)

Plug

x = 10

into

- 2 x^{3} + 42 x^{2} - 240 x + 5 : - 195

Maximum (10, - 195)

Minimum (4, - 411), Maximum (10, - 195)

y = 6 x - x^{2} - 2

f (x) = \frac{1}{( x ^{2} - 2 x + 7 )}

f (x) = 3 x^{2} - 27

x^{2} - 14 x + 3

f (x) = 39 + 6 x + \frac{54}{x}

What is the extreme f(x)=-2x^3+42x^2-240x+5 ?
The extreme f(x)=-2x^3+42x^2-240x+5 is Minimum(4,-411),Maximum(10,-195)