What is the extreme f(x)=(x^2-9x+39)/(x-7) ?

The extreme f(x)=(x^2-9x+39)/(x-7) is Maximum(2,-5),Minimum(12,15)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(x^2-9x+39)/(x-7)

extreme

Maximum (2, - 5), Minimum (12, 15)

Solution steps

f^{'} (x) = \frac{x ^{2} - 14 x + 24}{( x - 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 7,

Decreasing

: 7 < x < 12,

Increasing

: 12 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 9 x + 39}{x - 7} : - 5

Maximum (2, - 5)

Plug

x = 12

into

\frac{x ^{2} - 9 x + 39}{x - 7} : 15

Minimum (12, 15)

Maximum (2, - 5), Minimum (12, 15)

What is the extreme f(x)=(x^2-9x+39)/(x-7) ?
The extreme f(x)=(x^2-9x+39)/(x-7) is Maximum(2,-5),Minimum(12,15)