What is the extreme (x^2-7x+26)/(x-5) ?

The extreme (x^2-7x+26)/(x-5) is Maximum(1,-5),Minimum(9,11)

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extreme (x^2-7x+26)/(x-5)

extreme points

\frac{x ^{2} - 7 x + 26}{x - 5}

Maximum (1, - 5), Minimum (9, 11)

Solution steps

f^{'} (x) = \frac{x ^{2} - 10 x + 9}{( x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 5,

Decreasing

: 5 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 1

into

\frac{x ^{2} - 7 x + 26}{x - 5} : - 5

Maximum (1, - 5)

Plug

x = 9

into

\frac{x ^{2} - 7 x + 26}{x - 5} : 11

Minimum (9, 11)

Maximum (1, - 5), Minimum (9, 11)

f (x) = 3 x^{3} e^{- x}, - 1 \leq x \leq 5

(x + 8)^{\frac{2}{3}}

f (x) = x^{2} - y^{2} - 7 x - 4 y

f (x, y) = x^{3} + y^{3} - 21 x y

f (x) = - 3 x + ln (x)

What is the extreme (x^2-7x+26)/(x-5) ?
The extreme (x^2-7x+26)/(x-5) is Maximum(1,-5),Minimum(9,11)