Find the inverse function

[ryleyzip123]

Find the inverse function. f(x)=1+sqrt(x-5).

f^-1(x)=(x-?)^2+? for x>/=?

 

[Deleted member 4993]

Find the inverse function. f(x)=1+sqrt(x-5).

f^-1(x)=(x-?)^2+? for x>/=?

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about these problems.

 

[ryleyzip123]

when I find the inverse of this function, I get x^2-2x+6. I also have figured out that (x-1)^2+5=x^2-2x+6. When submitting that answer [(x-1)^2+5 for x>=0], it said the restriction of x is incorrect. The range of f is the domain of f^-1. I am stuck there

 

[Dr.Peterson]

when I find the inverse of this function, I get x^2-2x+6. I also have figured out that (x-1)^2+5=x^2-2x+6. When submitting that answer [(x-1)^2+5 for x>=0], it said the restriction of x is incorrect. The range of f is the domain of f^-1. I am stuck there

Your last comment is valuable. What is the range of f? If you look at the function, the range of the radical is [MATH]y\ge 0[/MATH], and when we add 1 to that, we get what?

That is, the range is determined by the fact that the radical is always non-negative. And this is what determines the domain of the inverse.

As for the inverse itself, you evidently went farther than you needed to when you expanded to [MATH]x^2-2x+6[/MATH]. In solving for it, you presumably wrote [MATH]x=1+\sqrt{y-5}[/MATH], then subtracted 1 to get [MATH]x-1=\sqrt{y-5}[/MATH] and squared to get [MATH](x-1)^2 = y-5[/MATH]. Seeing the form they want, I would not expand the left-hand side, but just add 5 to each side to get [MATH]y = (x-1)^2 + 5[/MATH].

 

[Deleted member 4993]

when I find the inverse of this function, I get x^2-2x+6. I also have figured out that (x-1)^2+5=x^2-2x+6. When submitting that answer [(x-1)^2+5 for x>=0], it said the restriction of x is incorrect. The range of f is the domain of f^-1. I am stuck there

What is the restriction 'y' for your original function?

Remember √ is always positive non-negative .

 

[ryleyzip123]

Your last comment is valuable. What is the range of f? If you look at the function, the range of the radical is [MATH]y\ge 0[/MATH], and when we add 1 to that, we get what?

That is, the range is determined by the fact that the radical is always non-negative. And this is what determines the domain of the inverse.

As for the inverse itself, you evidently went farther than you needed to when you expanded to [MATH]x^2-2x+6[/MATH]. In solving for it, you presumably wrote [MATH]x=1+\sqrt{y-5}[/MATH], then subtracted 1 to get [MATH]x-1=\sqrt{y-5}[/MATH] and squared to get [MATH](x-1)^2 = y-5[/MATH]. Seeing the form they want, I would not expand the left-hand side, but just add 5 to each side to get [MATH]y = (x-1)^2 + 5[/MATH].

So, would the correct answer be f^-1=(x-1)^2+5 for x>=1? because you add the 1 to the zero?

 

[Dr.Peterson]

Yes.

 

[lookagain]

Remember √ is always positive.

The value of a square root is non-negative.

 

[Deleted member 4993]

The value of a square root is non-negative.

Yes ... corrected.

 

[Steven G]

Just do things in reverse.

x -->x-5 --> sqrt(x-5) ---> sqrt(x-5) + 1
(x-1)^2 + 5 <-- (x-1)^2 <-- x-1 <-- x