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Question 455309: How do you solve domains of functions? For example,
f(x)=square root of (x-2)
f(x)=-5x/7+4
f(x)=1/x-2
There's the domain of f(x), the range of f(x), the equation of f^-1(x), the range of f^-1(x), and whether f^-1(x) is a function. How do I solve for each?
Thanks!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! How do you solve domains of functions? For example,
f(x)=square root of (x-2)
f(x)=-5x/7+4
f(x)=1/x-2
There's the domain of f(x), the range of f(x), the equation of f^-1(x), the range of f^-1(x), and whether f^-1(x) is a function. How do I solve for each?
..
f(x)=square root of (x-2)
When finding domains, look for restrictions. In this case it is a square root where the radican must be ≥0. To find the restriction, set the radican≥0, then solve the inequality.
x-2≥0
x≥2
Domain: [2,∞) This means only x-values greater or equal to 2 are valid for this function.
To find the range, interchange x and y, then solve for y.
y=sqrt(x-2)
x=sqrt(y-2)
x^2=y-2
y=x^2+2 (inverse)
Range:[0,∞) Note that y cannot be negative because of the x^2 term
Equation of f^-1(x), the inverse of f(x). You just found it, y=x^2+2. Use the same method for finding the range previously. The range and domain of the inverse function are interchanged with those of the original function,f(x). For the inverse to exist the original must be a one-to-one function, that is, each x-value can only have one y-value.
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f(x)=-5x/7+4
In this case, the restrictions are in the denominator. To find these restrictions, set the denominator=0, then solve for x. Your problem, however, shows no variables in the denominator so there are no restrictions for domain and range: (-∞,∞). You probably made a typing error.
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f(x)=1/x-2
In this case you do have a variable in the denominator.
set x-2=0
x=2 which means x≠2. When x=2, the denominator becomes 0, and f(x) becomes undefined.
Domain:(-∞,2) U (2,∞)
solving for range
y=1/x-2
x=1/y-2
xy-2x=1
y=(2x+1)/x (inverse)
From this you can see that x≠0, otherwise the function becomes undefined
Range:(-∞, 0) U (0, ∞)
The range and domain of the inverse function are interchanged with those of the original function,f(x).
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