What is the extreme f(x)=3x^2-30x+12,-10<= x<= 10 ?

The extreme f(x)=3x^2-30x+12,-10<= x<= 10 is Maximum(-10,612),Minimum(5,-63),Maximum(10,12)

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extreme f(x)=3x^2-30x+12,-10<= x<= 10

extreme points

f (x) = 3 x^{2} - 30 x + 12, - 10 \leq x \leq 10

Maximum (- 10, 612), Minimum (5, - 63), Maximum (10, 12)

Solution steps

Find the critical points:

x = - 10, x = 5, x = 10

Domain of

3 x^{2} - 30 x + 12, - 10 \leq x \leq 10 : - 10 \leq x \leq 10

Find intervals:

Decreasing

: - 10 < x < 5,

Increasing

: 5 < x < 10

Plug the extreme point

x = - 10

into

3 x^{2} - 30 x + 12 \Rightarrow y = 612

Maximum (- 10, 612)

Plug the extreme point

x = 5

into

3 x^{2} - 30 x + 12 \Rightarrow y = - 63

Minimum (5, - 63)

Plug the extreme point

x = 10

into

3 x^{2} - 30 x + 12 \Rightarrow y = 12

Maximum (10, 12)

Maximum (- 10, 612), Minimum (5, - 63), Maximum (10, 12)

\frac{x - 521}{x ^{2} + x - 132}

f (x) = 8 x^{3} - 7 ln (x^{3}), (0.12, 7)

3 x^{4} - 2

f (x) = \frac{27}{x ^{2} - 4 x - 5}

f (x) = \frac{1}{3} x^{3} - \frac{1}{2} x^{2} - 2 x = \frac{4}{3}

What is the extreme f(x)=3x^2-30x+12,-10<= x<= 10 ?
The extreme f(x)=3x^2-30x+12,-10<= x<= 10 is Maximum(-10,612),Minimum(5,-63),Maximum(10,12)