Why does the square root of an inverse function turn negative?

Question

For example,

$$f(x)=x^2$$

$$y=x^2$$

$$-\sqrt{x} = f^{-1}$$

Why does $\sqrt{x}$ become negative?

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Edit: Sorry for all the confusion, I will state the problem on my textbook and the solution.

"Find F^{-1} and verify that (f · f^{-1})(x) = (f^{-1} · f)(x) = x"

$$f(x) = x^{2}, x≤0$$

Solution:

$$ x=-\sqrt{y}$$

Interchange x and y.

$$ y=-\sqrt{x}$$ $$f^{-1} (x) = - \sqrt{x}$$

Verify.

For x≥0

$$(f^{-1} · f)(x) = f(-\sqrt{x}) = (\sqrt{x})^{2} = x$$

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And all I want to know is why $$y=x^{2}$$ become into $$f^{-1}(x)=-\sqrt.{x}$$

Answer 1

There seems to be a lot of confusion here. Let's remember this very important definition: a function maps any one input to only one output. This means that you can never have $f(x) = 3$ and $f(x) = 5$, because that would mean that $3=5$, which is clearly wrong. As a rule, we have that if $f(x) = a$ and $f(x) = b$ then $a = b$.

So let's take $f(x) = x^2$. You know what that looks like --- it's a parabola. For example, we have that $f(3) = 9$ and $f(-3) = 9$. This means that $f(x)$ is not invertible. Why? If you tried to invert this function, then $f^{-1}(9) = 3$ and $f^{-1}(9) = -3$. But $3 \neq -3$. So $f^{-1}$ violates the rule we laid out above.

Answer 2

The function $f(x)=x^2$ doesn't actually have an inverse function.

Let's see why:

If we follow the standard procedure for calculating the inverse function, we get that the inverse function should be $f^{-1}=\pm\sqrt{x}$.

Now, we have a problem. Back in precalculus, we were taught that a function must only have one output per input. But, $\pm\sqrt{x}$ gives two outputs!

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Edit

An answer to your edited problem. $y=x^2 \Rightarrow x=\pm\sqrt{y}$. On the left and the right side we take the square root. Don't forget the $\pm$!!