What is the extreme f(x)=(x-1)(x-6)^3+6,1<= x<= 4 ?

The extreme f(x)=(x-1)(x-6)^3+6,1<= x<= 4 is Maximum(1,6),Minimum(9/4 ,-15339/256),Maximum(4,-18)

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extreme f(x)=(x-1)(x-6)^3+6,1<= x<= 4

extreme points

f (x) = (x - 1) (x - 6)^{3} + 6, 1 \leq x \leq 4

Maximum (1, 6), Minimum (\frac{9}{4}, - \frac{15339}{256}), Maximum (4, - 18)

Solution steps

Find the critical points:

x = 1, x = \frac{9}{4}, x = 4

Domain of

(x - 1) (x - 6)^{3} + 6, 1 \leq x \leq 4 : 1 \leq x \leq 4

Find intervals:

Decreasing

: 1 < x < \frac{9}{4},

Increasing

: \frac{9}{4} < x < 4

Plug the extreme point

x = 1

into

(x - 1) (x - 6)^{3} + 6 \Rightarrow y = 6

Maximum (1, 6)

Plug the extreme point

x = \frac{9}{4}

into

(x - 1) (x - 6)^{3} + 6 \Rightarrow y = - \frac{15339}{256}

Minimum (\frac{9}{4}, - \frac{15339}{256})

Plug the extreme point

x = 4

into

(x - 1) (x - 6)^{3} + 6 \Rightarrow y = - 18

Maximum (4, - 18)

Maximum (1, 6), Minimum (\frac{9}{4}, - \frac{15339}{256}), Maximum (4, - 18)

f (x) = x^{4} - 5 x^{3} + x^{2} + 21 x - 18

f (x, y) = x^{2} - 2 x y + 2 y^{2} - 2 x + 2 y + 1

g (x, y) = e^{x} - x + e^{y} - y

\frac{1}{3 - 2 x - x ^{2}}

f (x) = x x^{2} + 36

What is the extreme f(x)=(x-1)(x-6)^3+6,1<= x<= 4 ?
The extreme f(x)=(x-1)(x-6)^3+6,1<= x<= 4 is Maximum(1,6),Minimum(9/4 ,-15339/256),Maximum(4,-18)