Give an example of a rational function (i. e., the quotient of two polynomials) f satisfying the following conditions: • f is not defined at 1. • f (-3) = 0. • f (3) = 9. • lim x→5 + f (x) = - [infinity] and lim x→5 - f (x) = [infinity]. Explain your reasoning.

Answer: 6x+18 / (-x²+6x-5)

Step-by-step explanation:

• f is not defined at 1

• f (-3) = 0

• f (3) = 9

• lim x→5 + f (x) = - [infinity]

• lim x→5 - f (x) = [infinity]

• f is not defined at 1

we need to have a denominator 0 for x = 1

so, 1 / (x-1)

• lim x→5 + f (x) = - [infinity]

• lim x→5 - f (x) = [infinity]

For the limits, we need to have

1 / (-x+5)

this way, lim x→5 + f (x) = - [infinity] and lim x→5 - f (x) = [infinity]

So far we have

1 / (x-1) (-x+5)

• f (-3) = 0

the nominator has to be 0 when x = - 3

this way, x+3

so, (x+3) / (x-1) (-x+5)

• f (3) = 9

All we need to do is multiply (x+3) / (x-1) (-x+5) by 6, so

6 (x+3) / (x-1) (-x+5) = 6x+18 / (-x²+6x-5)