What is the extreme f(x)=x^3-5x^2+3x+12,(-2,4) ?

The extreme f(x)=x^3-5x^2+3x+12,(-2,4) is Maximum(1/3 , 337/27),Minimum(3,3)

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extreme f(x)=x^3-5x^2+3x+12,(-2,4)

extreme points

f (x) = x^{3} - 5 x^{2} + 3 x + 12, (- 2, 4)

Maximum (\frac{1}{3}, \frac{337}{27}), Minimum (3, 3)

Solution steps

Find the critical points:

x = \frac{1}{3}, x = 3

Domain of

x^{3} - 5 x^{2} + 3 x + 12, (- 2, 4) : - 2 < x < 4

Find intervals:

Increasing

: - 2 < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < 3,

Increasing

: 3 < x < 4

Plug the extreme point

x = \frac{1}{3}

into

x^{3} - 5 x^{2} + 3 x + 12 \Rightarrow y = \frac{337}{27}

Maximum (\frac{1}{3}, \frac{337}{27})

Plug the extreme point

x = 3

into

x^{3} - 5 x^{2} + 3 x + 12 \Rightarrow y = 3

Minimum (3, 3)

Maximum (\frac{1}{3}, \frac{337}{27}), Minimum (3, 3)

f (x) = - 4 x^{2} + 16 x - 3, - 4 \leq x \leq 4

f (x, y) = - 294 \cdot x + 2 \cdot x^{3} + 6 \cdot x \cdot y^{2} - 3 \cdot y^{3}

f (x) = ln (x y - 6)

f (x) = e^{4 x} + e^{- x}

f (x, y) = 3 x^{2} + 2 y^{2} - 18 x + 16 y + 7

What is the extreme f(x)=x^3-5x^2+3x+12,(-2,4) ?
The extreme f(x)=x^3-5x^2+3x+12,(-2,4) is Maximum(1/3 , 337/27),Minimum(3,3)