What is the extreme g(x)=x^3-12x^2+45x+4 ?

The extreme g(x)=x^3-12x^2+45x+4 is Maximum(3,58),Minimum(5,54)

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extreme g(x)=x^3-12x^2+45x+4

extreme points

g (x) = x^{3} - 12 x^{2} + 45 x + 4

Maximum (3, 58), Minimum (5, 54)

Solution steps

f^{'} (x) = 3 x^{2} - 24 x + 45

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 3

into

x^{3} - 12 x^{2} + 45 x + 4 : 58

Maximum (3, 58)

Plug

x = 5

into

x^{3} - 12 x^{2} + 45 x + 4 : 54

Minimum (5, 54)

Maximum (3, 58), Minimum (5, 54)

g (x) = x^{3} - 12 x^{2} + 45 x + 1

g (x) = x^{3} - 12 x^{2} + 45 x + 2

g (x) = x^{3} - 12 x^{2} + 45 x + 7

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

f (x) = \frac{( x ^{2} - 1 )}{x ^{2} + 2 x - 3}

What is the extreme g(x)=x^3-12x^2+45x+4 ?
The extreme g(x)=x^3-12x^2+45x+4 is Maximum(3,58),Minimum(5,54)