Question

Find the domain of the composition (h(f(x))

F(x)= 3x^2-3, H(x)= ln 1/x-3
I used mathway to help get the answer which is (-Infinity symbol,-1) U (1,infinity symbol)

But I need to know the steps that was taken to get the answer

Answer 1

No I will Not find the domain of the composition >:( how DARE you tell me what to do ~samantha~ for I will SUE you. >:((((

Answer 2

F(x)= 3x^2-3,

H(x)= ln 1/x-3 = ln((x-3)^-1) = -ln(x-3)

h(f(x))
= h(3x^2 - 3) , assuming f(x) is the same as F(x), as well as the h(x) confusion
= -ln (3x^2 - 3 - 3)
= -ln(3x^2 - 6)

remember we can only take logs of positive numbers, so
3x^2 - 6 > 0
x^2 - 2 > 0
x^2 > 2

So x > 2 OR x < -2

BUT in h(x) = ln( 1/(x-3) ), we have a further implied restriction of x > 3
(For values of x ≤ 3 , we would first of all be dividing by zero in 1/(x-3)
and secondly we would be taking logs of a negative)

You have the interesting case where a value of x = 3 is undefined in one of the individual functions but is ok in the final composite function.

I would state the domain as x > 3
Check with your instructor how to deal with this case.

Answer 3

Thank you oobleck. I will, because I remember doing this before but forgot over time.

Answer 4

[AI-generated answer — Explain Bot]

To find the domain of the composition h(f(x)), we need to consider two things: the domain of the inner function f(x) and the domain of the outer function h(x). Let's break down the steps:

1. Start with the domain of the inner function f(x). In this case, we have f(x) = 3x^2 - 3.

Since f(x) is a quadratic function, it is defined for all real numbers. So its domain is (-∞, ∞).

2. Now, move on to the domain of the outer function h(x). In this case, we have h(x) = ln(1/x - 3).

Since we have a natural logarithm function, the argument (1/x - 3) must be greater than zero for the function to be defined. Therefore, we need to solve the inequality 1/x - 3 > 0.

To do this, we first find the critical point by setting the expression inside the logarithm equal to zero:

1/x - 3 = 0

Solving for x, we get:

1/x = 3

x = 1/3

Now, we can create a sign chart to determine when 1/x - 3 is positive or negative:

(-∞) | (1/3) | (∞)
------------------------------------
(+) (-)

From the sign chart, we see that 1/x - 3 is positive for x < 1/3, and negative for x > 1/3.

However, we also need to consider the excluded value x = 0 since division by zero is undefined. So the domain of h(x) is (-∞, 0) U (0, 1/3) U (1/3, ∞).

3. Finally, we need to consider the composition h(f(x)), where we substitute f(x) = 3x^2 - 3 into h(x).

We need to ensure that the result of f(x) falls within the domain of h(x).

Plugging f(x) into h(x), we get:

h(f(x)) = ln(1/(3x^2 - 3) - 3)

Since the expression 3x^2 - 3 does not equal to zero for any real value of x, we don't need to exclude any values from the domain of h(f(x)).

Thus, the domain of h(f(x)) is the same as the domain of h(x), which is (-∞, 0) U (0, 1/3) U (1/3, ∞).