Let f(x)=1/x and g(x)=x^2 + 5x.

a. Find (f*g)(x)
b. Find the domain and range of (f*g)(x).

I know how to set up the problem, but I don't know where to go from there.

(f*g)(x)=(1/x)(x^2+5x)

but that is

x+5
straight line

domain is all real numbers
range is all real numbers

So just to be sure...

(f*g)(x)=(1/x)*x(x+5) - factor the expression, (x^2+5x)
(f*g)(x)=(x(x+5))/x - multiply the two expressions
(f*g)(x)=x+5 - the x/x cancel out leaving x+5

Right?

Yes, but you sure made it complicated :)

I tend to do that, thank you!

The point is that even x^2/x

would have a zero in the denominator as x --->0
BUT
(x/x) x = x, like period. The top goes to zero at exactly the same rate as the bottom, so the expression is really 1 x

To find (f*g)(x), we need to find the composition of the two functions f(x) and g(x). The composition is denoted as (f*g)(x) and is found by evaluating g(x) and then substituting it into f(x).

a. Evaluating g(x):
g(x) = x^2 + 5x

Now we substitute g(x) into f(x):
(f*g)(x) = f(g(x)) = f(x^2 + 5x)

To find (f*g)(x), we substitute x^2 + 5x into f(x):
(f*g)(x) = 1 / (x^2 + 5x)

b. To find the domain and range of (f*g)(x), we need to determine the restrictions on x for which the expression is defined.

Domain: The function (f*g)(x) is defined for all real numbers except when the denominator (x^2 + 5x) equals zero, since we cannot divide by zero. So to find the domain, we need to solve the equation x^2 + 5x = 0.

Factoring the equation:
x(x + 5) = 0

Setting each factor equal to zero:
x = 0 or x + 5 = 0

Solving for x:
x = 0 or x = -5

Therefore, the domain of (f*g)(x) is all real numbers except x = 0 and x = -5.

Range: To find the range of (f*g)(x), we need to analyze the behavior of the function. As x approaches positive infinity or negative infinity, the expression 1 / (x^2 + 5x) approaches zero. Hence, the range of (f*g)(x) is all real numbers except zero.

In summary,
a. (f*g)(x) = 1 / (x^2 + 5x)
b. Domain is all real numbers except x = 0 and x = -5.
Range is all real numbers except zero.