What is the extreme f(x)=20x^3+9x^2-24x+12,0<= x<= 1 ?

The extreme f(x)=20x^3+9x^2-24x+12,0<= x<= 1 is Maximum(0,12),Minimum(1/2 , 19/4),Maximum(1,17)

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extreme f(x)=20x^3+9x^2-24x+12,0<= x<= 1

extreme points

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, 0 \leq x \leq 1

Maximum (0, 12), Minimum (\frac{1}{2}, \frac{19}{4}), Maximum (1, 17)

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}, x = 1

Domain of

20 x^{3} + 9 x^{2} - 24 x + 12, 0 \leq x \leq 1 : 0 \leq x \leq 1

Find intervals:

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < 1

Plug the extreme point

x = 0

into

20 x^{3} + 9 x^{2} - 24 x + 12 \Rightarrow y = 12

Maximum (0, 12)

Plug the extreme point

x = \frac{1}{2}

into

20 x^{3} + 9 x^{2} - 24 x + 12 \Rightarrow y = \frac{19}{4}

Minimum (\frac{1}{2}, \frac{19}{4})

Plug the extreme point

x = 1

into

20 x^{3} + 9 x^{2} - 24 x + 12 \Rightarrow y = 17

Maximum (1, 17)

Maximum (0, 12), Minimum (\frac{1}{2}, \frac{19}{4}), Maximum (1, 17)

5 x^{3} - 30 x^{2} + 45 x

f (x) = - x + 2 cos (x), 0 \leq x \leq 2 π

+ (x - 1)^{2}

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

y = sin (x) cos (x), 0 \leq x \leq π

What is the extreme f(x)=20x^3+9x^2-24x+12,0<= x<= 1 ?
The extreme f(x)=20x^3+9x^2-24x+12,0<= x<= 1 is Maximum(0,12),Minimum(1/2 , 19/4),Maximum(1,17)