What is the extreme (x^2-6x+12)/(x-4) ?

The extreme (x^2-6x+12)/(x-4) is Maximum(2,-2),Minimum(6,6)

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extreme (x^2-6x+12)/(x-4)

extreme points

\frac{x ^{2} - 6 x + 12}{x - 4}

Maximum (2, - 2), Minimum (6, 6)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 4,

Decreasing

: 4 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 6 x + 12}{x - 4} : - 2

Maximum (2, - 2)

Plug

x = 6

into

\frac{x ^{2} - 6 x + 12}{x - 4} : 6

Minimum (6, 6)

Maximum (2, - 2), Minimum (6, 6)

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

f (x) = (2 x^{2} - 2)^{2}

y = ∣ x ∣

f (x) = 2 x - 3 sin (x)

f (t) = t - 3 t, - 1 \leq t \leq 4

What is the extreme (x^2-6x+12)/(x-4) ?
The extreme (x^2-6x+12)/(x-4) is Maximum(2,-2),Minimum(6,6)