What is the extreme f(x)=2x^3-33x^2+144x+3,(4,9) ?

The extreme f(x)=2x^3-33x^2+144x+3,(4,9) is Minimum(8,67)

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extreme f(x)=2x^3-33x^2+144x+3,(4,9)

extreme points

f (x) = 2 x^{3} - 33 x^{2} + 144 x + 3, (4, 9)

Minimum (8, 67)

Solution steps

Find the critical points:

x = 8

Domain of

2 x^{3} - 33 x^{2} + 144 x + 3, (4, 9) : 4 < x < 9

Find intervals:

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < 9

Plug the extreme point

x = 8

into

2 x^{3} - 33 x^{2} + 144 x + 3 \Rightarrow y = 67

Minimum (8, 67)

4 - 5 x^{2}

h (x) = x^{3} + 12 x

f (x) = - 3 x^{2} + 2 y^{2} + 4 x y

f (x) = ln (x^{2} + 3 x + 9), - 2 \leq x \leq 2

\frac{1}{3 x + 6}

What is the extreme f(x)=2x^3-33x^2+144x+3,(4,9) ?
The extreme f(x)=2x^3-33x^2+144x+3,(4,9) is Minimum(8,67)