Explaining the Inverse of f(x)=sq.rt.(x+1) and f(x)=sq.rt.(x+10)

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In summary, the conversation discusses the inverse of two functions, f(x)=sqrt(x+1) and f(x)=sqrt(x+10). The question is raised about why the inverse must have the condition "x is greater than or equal to 0." The range of square roots is also discussed, with the conclusion that the range is [0, infinity). The concept of switching the x and y values to find the inverse is mentioned, along with a pattern seen in the graphs of y=sqrt(x+1) and y=x^2-1. Finally, it is explained that the domain and range are swapped when finding the inverse of a function.
  • #1
AznBoi
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Ok here are some examples that I am confused with:

f(x)=sq.rt.(x+1)
Inverse: y=x^2-1, x greater or equal to 0??

f(x)=sq.rt.(x+10)
Inverse: y=x^2-10, x greater or equal to 0??

How come you need the "x is greater or equal to 0" after each inverse?Can someone explain to me why this is? Is it because you can only inverse the positive side of the sq.rt.? How come you can't inverse the other side? I just don't get why you need to right that in the answer. Thanks a lot! :smile:
 
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  • #2
In your two expressions for the f's, what are their maximal range?
 
  • #3
The range of sq.rts. are infinity. or y is greater or equal to 0. Is that why?
 
  • #4
AznBoi said:
The range of sq.rts. are infinity. or y is greater or equal to 0. Is that why?
"The range of sqrt. are infinity", whatever does that mean?
 
  • #5
The range of square roots are (0, infinity) on the positive side. max range= infinity?
 
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  • #6
If you think of a function as a machine, where you put in an x, it does something to it, and shoots out a y, then all the possible y values make up the range. So you are right, the range of both is [tex]y\geq0[/tex] What happens to the range and domain to a function when it is converted to it's inverse? Take a look at the graphs it should become apparent. If not, look at the table of values.
 
  • #7
hint: If you were to switch all the x and y values of a function, what would happen? The points (0.1) (3,2) and (8,3) all lie on the graph of [tex]y=\sqrt{x+1}[/tex]. The points (1,0) (2,3) (3,8) all lie on the graph of [tex]y=x^2-1[/tex] See a pattern? What would that do to the domain and range?

Sorry for the premature post, I tried switching to advanced mode and accidentally pressed submit message.
 
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  • #8
AznBoi said:
The range of square roots are (0, infinity) on the positive side. max range= infinity?
That's much better than "are infinity"! Although I would say [0, infinity), specifically including 0 as a possible value. The point is that, if f is a function from A to B, then f-1 is a function from B to A: domain and range are swapped. If the range of f is [0, infinity), then the domain of f-1 is [0, infinity).
 
  • #9
thanks a lot :-p
 

Related to Explaining the Inverse of f(x)=sq.rt.(x+1) and f(x)=sq.rt.(x+10)

1. What is the inverse of f(x)=sq.rt.(x+1)?

The inverse of a function is a new function that undoes the original function. In this case, the inverse of f(x)=sq.rt.(x+1) is f(x)=x^2-1. This means that if you plug in the output of f(x)=sq.rt.(x+1) into the inverse function, you will get back the original input.

2. How do you find the inverse of a square root function?

To find the inverse of a square root function, switch the x and y variables and solve for y. This will give you the inverse function. In the case of f(x)=sq.rt.(x+1), switching the variables gives you x=sq.rt.(y+1). Now, solve for y by squaring both sides and subtracting 1, which gives you y=x^2-1. Therefore, the inverse of f(x)=sq.rt.(x+1) is f(x)=x^2-1.

3. What is the domain and range of the inverse of f(x)=sq.rt.(x+1)?

The domain of the inverse function is the range of the original function, which means all the possible inputs of the original function become the outputs of the inverse function. In this case, the domain of the inverse function is all real numbers. The range of the inverse function is the domain of the original function, which means all the possible outputs of the original function become the inputs of the inverse function. In this case, the range of the inverse function is all real numbers greater than or equal to -1.

4. How do you graph the inverse of f(x)=sq.rt.(x+1)?

To graph the inverse of a function, switch the x and y variables and plot the points to create a new graph. In this case, the original function f(x)=sq.rt.(x+1) has a square root graph shifted 1 unit to the left, so the inverse function will have a parabola graph shifted 1 unit down. You can also verify the graph by checking that the points (0,1) and (1,2) on the original graph will be switched to (1,0) and (2,-1) on the inverse graph.

5. Is the inverse of f(x)=sq.rt.(x+1) the same as the inverse of f(x)=sq.rt.(x+10)?

No, the inverse of f(x)=sq.rt.(x+1) is f(x)=x^2-1, while the inverse of f(x)=sq.rt.(x+10) is f(x)=x^2-10. The two inverse functions have different equations, and their graphs will also be different. The only similarity is that both inverse functions will undo the original function f(x)=sq.rt.(x+1) and f(x)=sq.rt.(x+10), respectively.

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