If f(x)= ln x, g(x) = e^3x, and h(x) = x^7, find the following...
a) (fog)(x) What is the domain of fog?
b) (gof)(x) What is the domain of gof?
c)(foh) What is the domain of foh?
a)
f(g) = ln(g) = ln(e^3x) = 3x
b)
g(f) = e^(3f) = e^(3lnx) = e^(ln(x^3)) = x^3
c)
f(h) = ln(h) = ln(x^7) = 7lnx
you can prolly work out the domains. Just remember that the domain of f and of g must be considered when deciding on the domain of (f◦g). If you have trouble, some on back and show where you get stuck.
To find the compositions of functions, we can substitute one function into the other and perform the necessary operations.
a) Let's start with (fog)(x), which means we substitute g(x) into f(x):
(fog)(x) = f(g(x)) = f(e^3x)
To find the domain of fog, we need to consider the domains of both functions f(x) and g(x).
The function f(x) = ln(x) is defined only for positive real numbers. So, the domain of f(x) is (0, +∞).
The function g(x) = e^3x is defined for all real numbers, as exponential functions are defined for any real value of x. Therefore, the domain of g(x) is (-∞, +∞).
Since we need to consider the composition of both functions, the domain of fog will be determined by the restrictions of both functions.
In this case, since g(x) is defined for all real numbers, the only domain restriction we need to consider is the domain of f(x).
Therefore, the domain of fog is (0, +∞).
b) Now, let's find (gof)(x), which means we substitute f(x) into g(x):
(gof)(x) = g(f(x)) = g(ln(x))
To find the domain of gof, we again need to consider the domains of f(x) and g(x).
The domain of f(x) is (0, +∞), as explained in part a). The domain of g(x) is (-∞, +∞), as exponential functions are defined for any real value of x.
Since we are composing the functions in the order gof, we need to consider the restrictions imposed by both functions.
In this case, the domain of gof will be determined by the restriction of g(x) since f(x) is defined for all real numbers within the domain of g(x).
Therefore, the domain of gof is (-∞, +∞).
c) Finally, let's calculate (foh)(x), which means we substitute h(x) into f(x):
(foh)(x) = f(h(x)) = f(x^7)
Once again, we need to consider the domains of both functions involved.
The domain of f(x) is (0, +∞) and the domain of h(x) = x^7 is (-∞, +∞) since it is defined for all real numbers.
Since we are composing the functions in the order foh, we need to consider the restrictions imposed by both functions.
In this case, the domain of foh will be determined by the restriction of f(x) since h(x) is defined for all real numbers within the domain of f(x).
Therefore, the domain of foh is (0, +∞).