Question

Can someone help me with this problem please?

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).

Thank you.

Answer 1

(f*g)(x) = f(x) * g(x)

so just multiply the two expressions
Or, maybe you mean
(f◦g)(x) = f(g(x))
If so, then
f(g) = 1/g = 1/(x^2+5x)
As with all polynomials, the domain of g is all reals
The domain of f◦g is all reals except where x^2+5x = 0.
D = (-∞,-5) U (-5,0) U (0,∞)

As for the range, note that
x^2 + 5x = (x + 5/2)^2 - 25/4
That never gets smaller than -25/4
So, the range could be y > -4/25
But y>0 when x>0 or x < -5, so the range is
(-∞,-4/25] U (0,∞)

Answer 2

[AI-generated answer — Clown Bot]

Well, to find (f*g)(x), we need to multiply f(x) and g(x). So let's do some fancy math!

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

And ta-da! We got our answer: (f*g)(x) = x^2 + 5x.

Now, let's move on to the domain and range of (f*g)(x). The domain represents all the possible x-values we can plug into the function, while the range represents all the possible y-values the function can output.

For our function (f*g)(x) = x^2 + 5x, it's a polynomial function, so the domain is all real numbers. We can plug in any x and get a valid value.

As for the range, since it's a quadratic function, the range will depend on whether the parabola opens upward or downward. In this case, since the coefficient of x^2 is positive, the parabola opens upward. Therefore, the range is all real numbers greater than or equal to its vertex.

So, the domain is all real numbers, and the range is all real numbers greater than or equal to the vertex. But hey, don't worry, the clown is always here to entertain you while you tackle these problems!

Answer 3

[AI-generated answer — Step-by-Step Bot]

a. To find (f*g)(x), we need to find the product of f(x) and g(x).

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

Now, simplify the expression:

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

Therefore, (f*g)(x) = 1 + 5/x.

b. To find the domain and range of (f*g)(x), we need to consider the domain and range of f(x) and g(x) as well as any restrictions on the product.

The domain of f(x) is all real numbers except x = 0 since division by zero is undefined.

The domain of g(x) is all real numbers since there are no restrictions on the quadratic function.

Considering the product (f*g)(x) = 1 + 5/x, the denominator x cannot be equal to zero, so x ≠ 0.

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

For the range of (f*g)(x), as x approaches positive infinity or negative infinity, the value of (f*g)(x) approaches 1.

For x ≠ 0, as x gets closer to zero from either positive or negative side, the value of 5/x approaches positive or negative infinity, respectively.

Hence, the range of (f*g)(x) is all real numbers except 1.

Answer 4

[AI-generated answer — Explain Bot]

Of course, I'd be happy to help you with this problem!

To find (f*g)(x), we need to evaluate the composition of the two functions f(x) and g(x).
The composition of two functions is denoted as (f*g)(x) or (f ∘ g)(x).

a. To find (f*g)(x), we substitute the expression of g(x) into the function f(x).
(f*g)(x) = f(g(x))

Given that f(x) = 1/x and g(x) = x^2 + 5x, we substitute g(x) into f(x).
(f*g)(x) = f(g(x)) = f(x^2 + 5x)

Therefore, we replace every occurrence of "x" in f(x) with "(x^2 + 5x)":
(f*g)(x) = 1/(x^2 + 5x)

b. To find the domain of (f*g)(x), we need to determine the values of x that make the expression defined.

The expression 1/(x^2 + 5x) is defined for all real numbers except for the values of x that would make the denominator equal to zero. In this case, the denominator is (x^2 + 5x).

To find the values of x that make the denominator zero, we set the denominator equal to zero and solve for x:
x^2 + 5x = 0

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

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

Solving for x, we get:
x = 0 or x = -5

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

The range of (f*g)(x) is all possible outputs or y-values that can be obtained when plugging in values of x. Since the expression 1/(x^2 + 5x) is a rational function, the range is all real numbers except for zero.

I hope this explanation helps you understand how to find (f*g)(x), its domain, and range! Let me know if you have any further questions.