Inverse functions

[jonnburton]

Hi,

I was wondering if anyone could tell me where I went wrong with trying to find the following inverse function:

\(\displaystyle h(x) = x^2 -3\)

so

\(\displaystyle y = x^2 - 3\)

adding 3 to both sides:

\(\displaystyle y +3 = x^2\)

\(\displaystyle x = \sqrt{y+3}\)

\(\displaystyle h^{-1}(x) = \sqrt{x+3}\)

However, the books answer says that \(\displaystyle h^{-1}(x) =- \sqrt{x+3}\)

 

[stapel]

....trying to find the following inverse function:

\(\displaystyle h(x) = x^2 -3\)

If they really gave you the above as the function for which you're supposed to find an inverse function, then "the answer" must be "there is no inverse function, as this original function does not pass the Horizontal Line Test".

Is there maybe more to this exercise, like a restricted domain on the original function? Because if there is, then graphing will make very obvious why there would be a restriction on the range of the answer.

By the way, when you learned about solving by taking square roots, didn't they tell you that a "plus/minus" sign went in front of the radical?

 

[Deleted member 4993]

Hi,

I was wondering if anyone could tell me where I went wrong with trying to find the following inverse function:

\(\displaystyle h(x) = x^2 -3\)

so

\(\displaystyle y = x^2 - 3\)

adding 3 to both sides:

\(\displaystyle y +3 = x^2\)

\(\displaystyle x = \sqrt{y+3}\)

\(\displaystyle h^{-1}(x) = \sqrt{x+3}\)

However, the books answer says that \(\displaystyle h^{-1}(x) =- \sqrt{x+3}\)

Was the domain and range of h(x) specified in the problem

Your inverse of the function should be \(\displaystyle h^{-1}(x) = \ \pm \sqrt{x+3}\)

If the given domain of h(x) was (-∞,0) then \(\displaystyle h^{-1}(x) =- \sqrt{x+3}\) would be correct [i.e. the range of h^-1(x) would be (-∞,0)]

 

[HallsofIvy]

As stapel said, the function you give does not have an inverse. You might recall the "vertical line" test for a function- f(x) is a function if and only if a vertical line (a fixed value of x in the domain of the function) crosses the graph of y= f(x) only once. That is each value of x gives only one value of y. Since changing to the inverse function "swaps" x and y, it also "swaps" horizontal and vertical. That is, a function, f(x), has an inverse if and only if a horizontal line crosses the graph only once. This is clearly NOT the case for \(\displaystyle y= x^2- 3\). The horizontal line at y= 1 crosses the graph at both (2, 1) and (-2, 1).

If we "cut" the graph so that each part satisifies the "horizontal line test", by restricting the domain, we can find an inverse for this new function. \(\displaystyle x^2- 3\) "turns" at x= 0 (the "vertex" is (0, -3)) so we cut there. That gives two functions: \(\displaystyle f_1(x)= x^2- 3\) for \(\displaystyle x\ge 0\) and \(\displaystyle f_2(x)= x^2- 3\) for \(\displaystyle x\le 0\).
Yes, solving for x we have \(\displaystyle x^2= y+ 3\) so that \(\displaystyle x= \pm\sqrt{y+ 3}\).

If the function was defined as "\(\displaystyle f(x)= x^2- 3\) for \(\displaystyle x\ge 0\)" then the inverse function would be \(\displaystyle f^{-1}= \sqrt{y+ 3}\). But if the function was defined as as "\(\displaystyle f(x)= x^2- 3\) for \(\displaystyle x\le 0\)" then the inverse function would be \(\displaystyle f^{-1}= -\sqrt{x+ 3}\).

Now, go back, look at the problem again, and see if the function wasn't actually defined as "\(\displaystyle f(x)= x^2- 3\) for \(\displaystyle x\le 0\)".

 

[jonnburton]

Thank you for your replies.

Yes, the function was defined for \(\displaystyle x\le 0\).

I see why the domain has to be restricted; I hadn't taken this into account properly before....

(And I had also forgotten that when solving square roots a plus/minus sign goes in front of the radical.)

 

[Deleted member 4993]

Thank you for your replies.

Yes, the function was defined for \(\displaystyle x\le 0\).

I see why the domain has to be restricted; I hadn't taken this into account properly before....

(And I had also forgotten that when solving square roots a plus/minus sign goes in front of the radical.)

But remember that when you put ± sign in front of an expression - it ceases to be a function. Functions must be single valued.

 

[lookagain]

Thank you for your replies.

Yes, the function was defined for \(\displaystyle x\le 0\).

The actual function isn't correctly given unless it is stated as the equivalent to: \(\displaystyle h(x) = x^2 - 3, \ \ x \le 0.\)