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
3 answers
No I will Not find the domain of the composition >:( how DARE you tell me what to do ~samantha~ for I will SUE you. >:((((
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.
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.
Thank you oobleck. I will, because I remember doing this before but forgot over time.