Question

Let f(x)= (x^2+3x-4) and g(x)= (x+4)

Answer 1

I. Multiply the first function by the second one.
f(x)*g(x) = (x^2+3x-4)*(x+4) = x^3 + 3x^2 - 4x + 4x^2 + 12x -16 = x^3 +7x^2 + 8x - 16.
The domain of this new function is the set of all real numbers (R). Other notation: from minus infinity to plus infinity. We came to this conclusion because the new function poses no restrictions; regardless of which x-value you take, you will get the appropriate y-value.

II. f(x)/g(x) = (x^2+3x-4)/(x+4) =
Ask yourself: which two numbers add up to 3 and multiply to -4? It's -1 and 4. Now we can represent f(x) as (x-1)(x+4).
Since we're dividing these 2 brackets by g(x)=x+4, we may now cancel (x+4). All that's left is x-1.
The domain here is the same as in the previous task - it is R.

Answer 2

Answer:

Ques 1)

$f(x)\cdot g(x)=x^3+7x^2+8x-16$

• The domain of f times g is all real numbers.

Ques 2)

$\dfrac{f}{g}=\dfrac{x^2+3x-4}{x+4}$

• All real numbers except x= -4

Step-by-step explanation:

We are given two functions f(x) and g(x) as follows:

$f(x)=x^2+3x-4$

and

$g(x)=x+4$

1)

$f\cdot g=f(x)\cdot g(x)\\\\i.e.\\\\f(x)\cdot g(x)=(x^2+3x-4)\cdot (x+4)\\\\i.e.\\\\f(x)\cdot g(x)=x^2(x+4)+3x(x+4)-4(x+4)\\\\i.e.\\\\f(x)\cdot g(x)=x^3+4x^2+3x^2+12x-4x-16\\\\i.e.\\\\f(x)\cdot g(x)=x^3+7x^2+8x-16$

Also, we know that the polynomial function is defined everywhere,.

i.e. the domain of the polynomial function is all real numbers.

Hence, the domain of f times g is all real numbers.

2)

$\dfrac{f}{g}=\dfrac{x^2+3x-4}{x+4}$

The function f/g is a rational function .

The domain of a rational function is all real numbers except the point where denominator is zero.

Hence, the domain of f/g is:

All real numbers except x= -4