What is the extreme f(x)=(x^2-16)/(x-5) ?

The extreme f(x)=(x^2-16)/(x-5) is Maximum(2,4),Minimum(8,16)

extreme f(x)=(x^2-16)/(x-5)

extreme points

f (x) = \frac{x ^{2} - 16}{x - 5}

Maximum (2, 4), Minimum (8, 16)

Solution steps

f^{'} (x) = \frac{x ^{2} - 10 x + 16}{( x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 5,

Decreasing

: 5 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 16}{x - 5} : 4

Maximum (2, 4)

Plug

x = 8

into

\frac{x ^{2} - 16}{x - 5} : 16

Minimum (8, 16)

Maximum (2, 4), Minimum (8, 16)

15 sin (5000 π t)

x^{4} + 2 x^{2} + y^{4} - 2 y^{2} + 3

f (x) = 9 \cdot x^{3} - 7 \cdot x^{2} + 3 \cdot x + 10

f (x) = x^{2} - 6, - 2 \leq x \leq 4

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

What is the extreme f(x)=(x^2-16)/(x-5) ?
The extreme f(x)=(x^2-16)/(x-5) is Maximum(2,4),Minimum(8,16)