What is the extreme f(x)=((x^2-8))/((x-3)) ?

The extreme f(x)=((x^2-8))/((x-3)) is Maximum(2,4),Minimum(4,8)

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extreme f(x)=((x^2-8))/((x-3))

extreme points

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

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

Solution steps

f^{'} (x) = \frac{x ^{2} - 6 x + 8}{( x - 3 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 3,

Decreasing

: 3 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 2

into

\frac{x ^{2} - 8}{x - 3} : 4

Maximum (2, 4)

Plug

x = 4

into

\frac{x ^{2} - 8}{x - 3} : 8

Minimum (4, 8)

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

f (x) = 4.3 x - 0.01 x^{2} - 8

f (x) = x^{5} - 3125 x, - 7 \leq x \leq 7

\frac{- 2 ln ( x ) - 2}{x ^{2}}

f (x) = x^{2} + \frac{200}{x}, [0, \infty]

3 x^{2} - 6 x - 9, 0 \leq x \leq 4

What is the extreme f(x)=((x^2-8))/((x-3)) ?
The extreme f(x)=((x^2-8))/((x-3)) is Maximum(2,4),Minimum(4,8)