Asked by Anonymous
f(x)=2x-4, g(x)=x squared + 5x
What is F o G and Domain
Okay, we are given f and g as functions of x. To compose a function means to use one function, g in this case, as the argument of the other, f in this case. Thus we want f o g = f(g(x))
With f(x) = 2x-4 and g(x) = x^2 + 5x, f o g becomes f(x^2 +5x)= 2(x^2 +5x) - 4.
What we do is substitue the equation for g into each place the argument of f occurs. I'll let you determine the domain.
Incidentally, it might be a little easier to grasp composite functions if you don't use the same variable in both functions. Example: suppose f= 2x-4, as you have above and g is a function of t, say g(t)=3t-1. Then in this case f o g = f(g(t)). To evaluate this, we put g(t) in place of the x in the equation 2x-4, so f o g = 2g(t)-4. We now substitute the equation for g(t) to get f o g = 2(3t-1)-4, which we would simplify to f(g(t))=6(t-1) for this example.
What is F o G and Domain
Okay, we are given f and g as functions of x. To compose a function means to use one function, g in this case, as the argument of the other, f in this case. Thus we want f o g = f(g(x))
With f(x) = 2x-4 and g(x) = x^2 + 5x, f o g becomes f(x^2 +5x)= 2(x^2 +5x) - 4.
What we do is substitue the equation for g into each place the argument of f occurs. I'll let you determine the domain.
Incidentally, it might be a little easier to grasp composite functions if you don't use the same variable in both functions. Example: suppose f= 2x-4, as you have above and g is a function of t, say g(t)=3t-1. Then in this case f o g = f(g(t)). To evaluate this, we put g(t) in place of the x in the equation 2x-4, so f o g = 2g(t)-4. We now substitute the equation for g(t) to get f o g = 2(3t-1)-4, which we would simplify to f(g(t))=6(t-1) for this example.
Answers
Answered by
JOe
Since the the domain of the inner function belongs to real numbers(a polynomial) and the new function is also a polynomial...therefore the domain is all real numbers.
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