Question

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 1

Answer:

$\frac{6x+18}{(x-1)(5-x)}$

Step-by-step explanation:

We have to find an example of  a rational function.

Where f is satisfying the following conditions

1.f is not defined at 1.

2.f(-3)=0

3.f(3)=9

4.$\lim_{x\rightarrow 5} f(x)=-\infty$

5.$\lim_{x\rightarrow}-f(x)=\infty$

If f is not defined at 1

f has (x-1) in the denominator

$\frac{1}{x-1}$

If $\lim_{x\rightarrow}f(x)=-\infty$

It means $\frac{1}{x-5}$

f has (5-x) in the denominator because $\frac{1}{0}=\infty$

$(x-1)(5-x)=-x^2+6x-5$

f(-3)=0

If f(-3) is zero it means numerator becomes zero when substitute x=-3 in the function

It means f has (x+3) in the numerator

f(3)=9

It means when multiply (x+3) by 6 then we get 9

because when x=3 then denominator

(3-1)(5-3)=4

When numerator is 6(x+3)

Then , substitute x=3

Then , numerator =36

After , dividing by 4 then we get 9

Therefore, we get f(3)=9

Hence,Rational function=$\frac{6x+18}{(x-1)(5-x)}$