What is the extreme f(x)=(x^2-35)/(x-6) ?

The extreme f(x)=(x^2-35)/(x-6) is Maximum(5,10),Minimum(7,14)

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extreme f(x)=(x^2-35)/(x-6)

extreme points

f (x) = \frac{x ^{2} - 35}{x - 6}

Maximum (5, 10), Minimum (7, 14)

Solution steps

f^{'} (x) = \frac{x ^{2} - 12 x + 35}{( x - 6 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 5,

Decreasing

: 5 < x < 6,

Decreasing

: 6 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 5

into

\frac{x ^{2} - 35}{x - 6} : 10

Maximum (5, 10)

Plug

x = 7

into

\frac{x ^{2} - 35}{x - 6} : 14

Minimum (7, 14)

Maximum (5, 10), Minimum (7, 14)

x^{3} - 18 x^{2} + 81 x

f (x, y) = x^{2} + x y + y^{2} + 8 y

f (x, y) = x^{2} + x y + y^{2} + 5 y

\frac{x ^{2} - 9}{x ^{2} - 4}

y = 2 x^{3} - 6 x + 1

What is the extreme f(x)=(x^2-35)/(x-6) ?
The extreme f(x)=(x^2-35)/(x-6) is Maximum(5,10),Minimum(7,14)