What is the extreme f(x)=-x^3+3x^2+45x-42 ?

The extreme f(x)=-x^3+3x^2+45x-42 is Minimum(-3,-123),Maximum(5,133)

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extreme f(x)=-x^3+3x^2+45x-42

extreme points

f (x) = - x^{3} + 3 x^{2} + 45 x - 42

Minimum (- 3, - 123), Maximum (5, 133)

Solution steps

f^{'} (x) = - 3 x^{2} + 6 x + 45

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = - 3

into

- x^{3} + 3 x^{2} + 45 x - 42 : - 123

Minimum (- 3, - 123)

Plug

x = 5

into

- x^{3} + 3 x^{2} + 45 x - 42 : 133

Maximum (5, 133)

Minimum (- 3, - 123), Maximum (5, 133)

f (x) = (x^{2} + 12) (1 - x^{2})

\frac{x ^{2}}{x - 2}

f (x) = \frac{- 3}{x ^{2} - 9}

f (x, y) = x y + \frac{9}{x} + \frac{3}{y}

6 x - x^{2}

What is the extreme f(x)=-x^3+3x^2+45x-42 ?
The extreme f(x)=-x^3+3x^2+45x-42 is Minimum(-3,-123),Maximum(5,133)