if f(x)=lnx, g(x)=e^3x, and h(x)=x^4 find following
A)(f of g)(x)and the domain of f of g
B) (g of f)(x) and the domain of g of f
C) (f of h)(x) and the domain of f of h
please show work
I will do B), you do the others the same way ...
b)
(g of f)(x) to me means g(f(x))
= g(lnx)
= e^(3lnx)
since we can only take logs of positive numbers,
domain: x > 0
range: I hope you are familiar with the basic exponential function, so y>0
To find the composite functions, we need to substitute one function into another.
A) (f of g)(x) means we substitute g(x) into f(x), which gives f(g(x)).
(f of g)(x) = f(g(x)) = ln(g(x))
Substituting g(x) = e^(3x) into f(x), we get:
(f of g)(x) = ln(g(x)) = ln(e^(3x))
Using the property that ln(e^x) = x, we simplify:
(f of g)(x) = ln(e^(3x)) = 3x
The domain of f(g(x)) is the same as the domain of g(x), which is all real numbers.
Domain of (f of g) = Domain of g = (-∞, ∞)
B) (g of f)(x) means we substitute f(x) into g(x), which gives g(f(x)).
(g of f)(x) = g(f(x)) = e^(3f(x))
Substituting f(x) = ln(x) into g(x), we get:
(g of f)(x) = e^(3f(x)) = e^(3ln(x))
Using the property e^(ln(x)) = x, we simplify:
(g of f)(x) = e^(3ln(x)) = x^3
The domain of g(f(x)) is the same as the domain of f(x), which is x>0 as ln(x) is only defined for x>0.
Domain of (g of f) = Domain of f = (0, ∞)
C) (f of h)(x) means we substitute h(x) into f(x), which gives f(h(x)).
(f of h)(x) = f(h(x)) = ln(h(x))
Substituting h(x) = x^4 into f(x), we get:
(f of h)(x) = ln(h(x)) = ln(x^4)
Using the property ln(a^b) = b ln(a), we simplify:
(f of h)(x) = ln(x^4) = 4 ln(x)
The domain of f(h(x)) is the same as the domain of h(x), which is all real numbers.
Domain of (f of h) = Domain of h = (-∞, ∞)
To find the composition of two functions, we substitute the expression for one function into the other function. Let's solve each part step by step:
A) (f o g)(x) and the domain of (f o g)
To find (f o g)(x), we substitute g(x) into f(x):
(f o g)(x) = f(g(x)) = ln(g(x))
Given that g(x) = e^(3x), we substitute it into f(x):
(f o g)(x) = ln(e^(3x))
We can simplify this by using a logarithmic property: ln(e^a) = a
(f o g)(x) = 3x
The domain of (f o g) is the same as the domain of g(x), which is all real numbers.
B) (g o f)(x) and the domain of (g o f)
To find (g o f)(x), we substitute f(x) into g(x):
(g o f)(x) = g(f(x)) = e^(3f(x))
Given that f(x) = ln(x), we substitute it into g(x):
(g o f)(x) = e^(3ln(x))
We can simplify this expression using the logarithmic property: e^(ln(a)) = a
(g o f)(x) = x^3
The domain of (g o f) is the same as the domain of f(x), which is x > 0 (positive real numbers).
C) (f o h)(x) and the domain of (f o h)
To find (f o h)(x), we substitute h(x) into f(x):
(f o h)(x) = f(h(x)) = ln(h(x))
Given that h(x) = x^4, we substitute it into f(x):
(f o h)(x) = ln(x^4)
Using a logarithmic property, ln(a^b) = bln(a):
(f o h)(x) = 4ln(x)
The domain of (f o h) is the same as the domain of h(x), which is all real numbers.