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
A) f(g(x))
=lnx(e^3x)
domain x>0
C)f(h(x))
=lnx(x^4)
x>4
A) close , but not quite
f(g(x))
= f(e^3x)
= ln (e^(3x)) , you have an extra x in there
f(x) was defined as lnx, so whatever x is replaced by on the left side, must be replaced by on the right side ....e.g.
f(2) = ln2
f(happy face) = ln(happy face)
f(e^3x) = ln(e^3x)
C) you made the same error
f(h(x)) = f(x^4) = ln x^4
the domain would be x > 0
i understand now thank you
B) (g of f)(x)
=(e^3x)^4
=(e^(3x*4))
=(e^(12x))
domain: -infinity < x < infinity
To find the composition of functions, we need to substitute the inner function into the outer function. Let's start with (f of g)(x), (g of f)(x), and (f of h)(x) one by one.
A) (f of g)(x):
To find (f of g)(x), we substitute g(x) into f(x):
(f of g)(x) = f(g(x)) = f(e^3x) = ln(e^3x) = 3x
The domain of (f of g)(x) is the same as the domain of g(x), which is all real numbers since e^3x is defined for all x.
B) (g of f)(x):
To find (g of f)(x), we substitute f(x) into g(x):
(g of f)(x) = g(f(x)) = g(lnx) = e^(3lnx)
To simplify this expression, we can use the property of logarithms:
e^(3lnx) = (e^lnx)^3 = x^3
The domain of (g of f)(x) is the same as the domain of f(x), which is x > 0 since lnx is defined only for positive values of x.
C) (f of h)(x):
To find (f of h)(x), we substitute h(x) into f(x):
(f of h)(x) = f(h(x)) = f(x^4) = ln(x^4) = 4lnx
The domain of (f of h)(x) is the same as the domain of h(x), which can be any real number since x^4 is defined for all x.