# Thread: Showing a function does not have an inverse

1. ## Showing a function does not have an inverse

Hello! I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. Please teach me how to do so using the example below! Thank you!

The function f is defined as f(x) = x^2 -2x -1, x is a real number. Show that the inverse function of f does not exist.

Thanks!

2. Originally Posted by Tangera
Hello! I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. Please teach me how to do so using the example below! Thank you!

The function f is defined as f(x) = x^2 -2x -1, x is a real number. Show that the inverse function of f does not exist.

Thanks!
I would show it using the quadratic formula.
Set the equation equal to zero:
$\displaystyle f(x)=x^2-2x-1=0$

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4*a*c}}{2a}$

Plug in values:
$\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4*(1)*(-1)}}{2(1)}$

Simplify:
$\displaystyle x=\frac{2\pm\sqrt{4+4}}{2}$

Simplify:
$\displaystyle x=\frac{2\pm\sqrt{8}}{2}$

Rewrite:
$\displaystyle x=\frac{2\pm 2\sqrt{2}}{2}$

Simplify:
$\displaystyle x=1\pm \sqrt{2}$

So, the inverse of f(x) when x=0 would return both $\displaystyle 1+\sqrt{2} \mbox{ and } 1-\sqrt{2}$ Thus it is not one-to-one.

Injective function - Wikipedia, the free encyclopedia

now you can show for example that this quadratic polynomial has two distinct roots:

$\displaystyle \Delta = \sqrt {( - 2)^2 - 4( - 1)} = \sqrt 8 > 0$

thus:

$\displaystyle f(x_1 ) = f(x_2 ) = 0\quad x_1 \ne x_2$

so the function is not 1:1

4. Well, using only f(x) = 0 is not enough for every quadratic. You can use a plotter and look for x values that give the same f(x). Or in this situation you can,

$\displaystyle f(x) = x^2 - 2x - 1$

$\displaystyle x^2 - 2x - 1 = a$

$\displaystyle x^2 - 2x + (-1-a) = 0$

$\displaystyle \Delta = b^2 - 4ac = 4 - (-4)(1)(-1-a)$

$\displaystyle \Delta = 8 + 4a$

For $\displaystyle a < -2,$ $\displaystyle \Delta < 0$ and it doesn't have a real root.
For $\displaystyle a = -2,$ $\displaystyle \Delta = 0$ and it has 2 equivalent roots. (or you can say that it has only one root)
For $\displaystyle a > -2,$ $\displaystyle \Delta > 0$ and it has 2 different real roots.

Thus, any a value greater than -2 will make $\displaystyle f^{-1}(a)$ have 2 values.

5. Originally Posted by Tangera
Hello! I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. Please teach me how to do so using the example below! Thank you!

The function f is defined as f(x) = x^2 -2x -1, x is a real number. Show that the inverse function of f does not exist.

Thanks!
Draw the graph of the function. Note that all functions satisfy the vertical line test - you can run a vertical line from left to right across the function and the vertical line will never cut more than once.

A function is 1-to-1 if it also satisfies the horizontal line test - you can run a horizontal line from down to up and the horizontal line will never cut the fuction more than once.

For your example, f(x) = x^2 -2x -1, run a horizontal line from down to up. After the horizontal line gets past the turning point, it cuts the parabola TWICE - that's more than once!! So f(x) = x^2 -2x -1 is NOT a 1-to-1 function.

If you restricted the domain of f(x) = x^2 -2x -1 = (x - 1)^2 - 2 to be x > 1, say, then the horizontal line will never cut more than once and so the function is now 1-to-1. Can you think of another restriction on the domain to make f(x) = x^2 -2x -1 a 1-to-1 function?

6. Originally Posted by Tangera
Hello! I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. Please teach me how to do so using the example below! Thank you!

The function f is defined as f(x) = x^2 -2x -1, x is a real number. Show that the inverse function of f does not exist.

Thanks!
Hi,

if you are allowed to use drivative you can use the property that with a one-to-one-function the sign of the first derivative doesn't change and the first derivative is unequal zero.

$\displaystyle f(x) = x^2 -2x -1$

$\displaystyle f'(x) = 2x-2$ has a change of sign from - to + at x = 1

and therefore f can't be a one-to-one-function.

7. Originally Posted by earboth
Hi,

if you are allowed to use drivative you can use the property that with a one-to-one-function the sign of the first derivative doesn't change and the first derivative is unequal zero.

$\displaystyle f(x) = x^2 -2x -1$

$\displaystyle f'(x) = 2x-2$ has a change of sign from - to + at x = 1

and therefore f can't be a one-to-one-function.
Wow, I never knew that! That's probably why I got docked points on one of my tests where we were supposed to show that the function only crossed the x-axis at one point. I put that since all the coefficients were positive, and all the powers of the derivatives were even, that it could never go from increasing to decreasing, then showed a point above and below the x-axis to show that it crossed at least once. I was kind of annoyed when I lost points on it, but I bet I was supposed to have used this property!

8. Erm.. sorry earboth, could you explain your derivative method again? I don't quite understand how/why it works. Thank you!

9. Originally Posted by Tangera
Erm.. sorry earboth, could you explain your derivative method again? I don't quite understand how/why it works. Thank you!
If f is a function which monotone (or monotonously or monotonically(?)) increases it satisfies the horizontal line test, the gradient of this function is positive and this function has an inverse function which increases monotone (or monotonously or monotonically(?)) too.
Take as an eaxample the function $\displaystyle f(x) = e^x$ and it's inverse function $\displaystyle f^{-1}(x) = \ln(x)$.

If the "behaviour" of a function changes from decreasing to increasing (example $\displaystyle p(x) = x^2$ then the sign of the gradient changes from - to +. With a differentiable function the first derivative will be zero at the point where the monotony changes. This point (turning point?) is a minimum point (in this case).

The function $\displaystyle f(x) = |x| + 1$ isn't differentiable at x = 0. But the gradient changes from -1 to +1 at x = 0 and therefore you have a minimum point at x =0.

If the monotony changes the function has at least 2 "branches" which will be intersected by a horizontal line twice. Such a function has no inverse function because the x-coordinates of the intersection points must be the y-coordinates of the inverse function and that's opposing the definition of a function.

10. Originally Posted by mr fantastic
If you restricted the domain of f(x) = x^2 -2x -1 = (x - 1)^2 - 2 to be x > 1, say, then the horizontal line will never cut more than once and so the function is now 1-to-1. Can you think of another restriction on the domain to make f(x) = x^2 -2x -1 a 1-to-1 function?
Another restriction could be x<1 or (this is random: ) 2<x<8 etc. Am I correct?

And thank you for explaining, Earboth!

11. Originally Posted by Tangera
Another restriction could be x<1 or (this is random: ) 2<x<8 etc. Am I correct? Mr F says: Totally

And thank you for explaining, Earboth!
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