If f(x)=x62 and g(x)= sqrt x-3 find fog and its domain.
To find the composition of the two functions, f(g(x)), we need to substitute g(x) into f(x).
Given that g(x) = sqrt(x-3), we have:
f(g(x)) = f(sqrt(x-3))
Now, substitute this function into f(x):
f(g(x)) = (sqrt(x-3))^2
Simplifying further:
f(g(x)) = (x-3)
Therefore, the composition of f(x) and g(x) is f(g(x)) = x-3.
The domain of f(g(x)) is the same as the domain of g(x) because that's the input used. In this case, the domain of g(x) is restricted by the square root function, which requires the expression inside the square root to be greater than or equal to zero.
So, x - 3 ≥ 0
=> x ≥ 3
Therefore, the domain of f(g(x)) is x ≥ 3.
To find the composition of two functions, f(x) and g(x), denoted as fog(x), you need to substitute the expression for g(x) into the expression for f(x) and simplify.
In this case, f(x) = x^2 and g(x) = sqrt(x-3).
To find fog(x), substitute g(x) into f(x):
fog(x) = f(g(x)) = f(sqrt(x-3))
Now, replace the x in f(x) with g(x) = sqrt(x-3):
fog(x) = (sqrt(x-3))^2
Simplifying this expression, we have:
fog(x) = x - 3
So, fog(x) = x - 3.
Now, let's determine the domain of fog(x).
The domain of a composition of functions is determined by the domain of the inner function (g(x)) that doesn't restrict the domain and the domain of the composition of functions (fog(x)) itself.
In this case, the inner function is g(x) = sqrt(x-3).
The domain of g(x) is the set of all real numbers that make the expression inside the square root non-negative. In other words, x - 3 ≥ 0.
Solving for x, we get x ≥ 3.
Therefore, the domain of g(x) is x ≥ 3.
Now, considering the composition fog(x) = x - 3, since there are no restrictions on the domain of fog(x), the domain of fog(x) is also x ≥ 3.
So, the domain of fog(x) is x ≥ 3.
f(g)=(sqrt(x-3))^2=x-3 or
x=f(g)+3
fog=x+3
(fog)(x) = f(g(x)) = [g(x)]^2
(fog)(x) = [sqrt(x-3)]^2
note that the sqrt and ^2 will just cancel each other out:
(fog)(x) = x - 3
to find its domain,, recall that domain is the set of all possible values of x,, since we both considered f(x) and g(x) here, we find the intersection of their domains.
for f(x) = x^2 , all real numbers are possible values of x
for g(x) = sqrt(x-3) , note that numbers less than 3 cannot be substituted here because the expression will become imaginary, for example if we take x = 2
g(-2) = sqrt(2-3) = sqrt (-1) = imaginary / does not exist
thus domain of g(x) is all numbers greater than or equal to 3
the intersection of the domains is thus
all numbers greater than or equal to 3 or in symbols,
[3, +infinity)
hope this helps~ :)