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

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

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

extreme points

f (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)

f (x) = x^{3} - 12 x^{2} + 45 x + 6

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

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

f (x) = x + ln (x^{2} + 1)

f (x, y) = \frac{x ^{2} + y ^{2} - 16}{x - 3}

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