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Eeshaan · Answer 1 · 2023-12-20T05:31:54+0000

Hello there. To solve this question, we'll have to remember some properties about functions.

Given the functions:

$\begin{gathered} f(x)=x^3 \\ g(x)=6x^2+11x-2 \end{gathered}$

We have to determine:

$\begin{gathered} (f+g)(x) \\ (f-g)(x) \\ (fg)(x) \\ (ff)(x) \\ \left((f)/(g)\right)(x) \\ \left((g)/(f)\right)(x) \end{gathered}$

And their domain.

Let's do each separately:

(f + g)(x)

In this case, this function is the same as adding f(x) and g(x):

And as it is a polynomial function, it has no holes or asymptotes, therefore its domain is all the real line. We write:

(f - g)(x)

In the same sense, it is equal to the difference between f and g:

(f-g)(x)=f(x)-g(x)=x^3-(6x^2+11x-2)=x^3-6x^2-11x+2

Again, as it is a polynomial function, its domain is all the real line, just as before.

(fg)(x)

In this case, it is the same as the product of f and g:

$(fg)(x)=f(x)\cdot g(x)=x^3\cdot(6x^2+11x-2)=6x^5+11x^4-2x^3$

Once again, its domain is all the real line.

(ff)(x)

In this case, it is the product of f and itself:

$(ff)(x)=f(x)\cdot f(x)=x^3\cdot x^3=x^6$

As before, its domain is entire real line.

(f/g)(x)

In this case, it is the quotient between f and g, respectively:

$\mleft((f)/(g)\mright)(x)=(x^3)/(6x^2+11x-2)$

But in this case, its domain is not the entire real line. We have to get rid of the holes and vertical asymptotes of the function.

This function has no holes, since we cannot simplify any terms in the fraction, but it has at least two vertical asymptotes (that we'll find by taking the roots of the denominator).

In fact, the name vertical asymptote stands for the values of x in which the function would not exist (its limit goes to either infinity, -infinity or would not exist).

These roots are given by:

Using the quadratic formula, we get:

$\begin{gathered} x=\frac{-11\pm\sqrt[]{11^2-4\cdot6\cdot(-2)}}{2\cdot6}=\frac{-11\pm\sqrt[]{121+48}}{12}=\frac{-11\pm\sqrt[]{169}}{12} \\ \Rightarrow x=(-11\pm13)/(12) \\ \Rightarrow x_1=(-11+13)/(12)=(2)/(12)=(1)/(6) \\ x_2=(-11-13)/(12)=(-24)/(12)=-2 \end{gathered}$

The roots are 1/6 and -2. They are the vertical asymptotes of the function.

The domain of (f/g)(x) is then given by subtracting these values from the real line:

Or also in interval notation:

We do the same to (g/f)(x):

It is equal to the quotient between g and f, respectively, thus

$\left((g)/(f)\right)(x)=(g(x))/(f(x))=(6x^2+11x-2)/(x^3)$

And again in this case, we have no holes, but we do have a vertical asymptote.

Taking the roots of the denominator: