What is the extreme f(x)=(5(x-6)^2)/(x-1) ?

The extreme f(x)=(5(x-6)^2)/(x-1) is Maximum(-4,-100),Minimum(6,0)

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extreme f(x)=(5(x-6)^2)/(x-1)

extreme points

f (x) = \frac{5 ( x - 6 ) ^{2}}{x - 1}

Maximum (- 4, - 100), Minimum (6, 0)

Solution steps

f^{'} (x) = \frac{5 ( x ^{2} - 2 x - 24 )}{( x - 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 1,

Decreasing

: 1 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - 4

into

\frac{5 ( x - 6 ) ^{2}}{x - 1} : - 100

Maximum (- 4, - 100)

Plug

x = 6

into

\frac{5 ( x - 6 ) ^{2}}{x - 1} : 0

Minimum (6, 0)

Maximum (- 4, - 100), Minimum (6, 0)

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

f (x) = x^{3} - x^{2} - 8 x + 12

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

f (x) = 5 x^{5} - 3 x^{2} + x - 10, - 5 \leq x \leq 6

f (x, y) = 32 y^{2} + x^{2} - x^{2} y

What is the extreme f(x)=(5(x-6)^2)/(x-1) ?
The extreme f(x)=(5(x-6)^2)/(x-1) is Maximum(-4,-100),Minimum(6,0)