Function Question

[BobBali]

Hi All,

question asks to show that the inverse of

$\displaystyle f(x)= x^2 - 2x + 5 is f^{-1} (x) = \sqrt{x-4} + 1$

However, this is a parabola to begin with, and does not satisfy the 'Horizontal line test' for whether a function has an inverse or not...

But anyways, i tried to find the inverse algebraically:

$\displaystyle x = y^2 - 2x +5$

$\displaystyle y^2 - 2y = x - 5$

$\displaystyle y(y-2) = x - 5$

$\displaystyle y = \frac{x-5}{y-2}$ *Not the ans as shown above...?

 

[pappus]

Hi All,

question asks to show that the inverse of

$\displaystyle f(x)= x^2 - 2x + 5 is f^{-1} (x) = \sqrt{x-4} + 1$<--- this is only true for $\displaystyle x \ge 4$

However, this is a parabola to begin with, and does not satisfy the 'Horizontal line test' for whether a function has an inverse or not... <--- therefore you must restrict the domain of the function such that it is monotonely increasing (x >=1) or monotonely decreasing (x<1)

But anyways, i tried to find the inverse algebraically:

$\displaystyle x = y^2 - 2x +5$<--- typo

$\displaystyle y^2 - 2y = x - 5$<--- miraculously correct

$\displaystyle y(y-2) = x - 5$

$\displaystyle y = \frac{x-5}{y-2}$ *Not the ans as shown above...?

1. Your equation is a quadratic equation in y:

$\displaystyle y^2 - 2y = x - 5~\implies~y^2-2y-(x-5)=0$

Apply the formula to solve a quadratic equation considering (x-5) as the constant summand:

2.

$\displaystyle y = \frac{2 \pm \sqrt{2^2-4 \cdot (-(x-5))}}{2}$

$\displaystyle y = 1 \pm \sqrt{1+(x-5)}$

$\displaystyle y = 1 \pm \sqrt{x-4}$

 

[lookagain]

$\displaystyle y = 1 \pm \sqrt{x-4}$

But then the solution is not done, because you would have to know/figure out
which branch to use, because it would be the right-half of the parabola
given first, to the problem solver and then the inverse function
(the upper or lower half) determined second/at the end.

Because, numerically, the range of the inverse function is the same as the
domain of the given function, look at the domain of the given function.
It was only looking at the inverse given by the OP that the domain of the
given function had to be [1, oo), but that is backwards.

Anyway, then the range of the inverse function must also be [1,oo).


So the inverse function must be $\displaystyle f^{-1}(x) = \sqrt{x - 4} \ + \ 1 \ \ (or \ \ 1 + \sqrt{x - 4}),$


which is the upper half of $\displaystyle f^{-1}(x) = 1 \pm \sqrt{x - 4}.$

 

[BobBali]

1. Your equation is a quadratic equation in y:

$\displaystyle y^2 - 2y = x - 5~\implies~y^2-2y-(x-5)=0$

Apply the formula to solve a quadratic equation considering (x-5) as the constant summand:

2.

$\displaystyle y = \frac{2 \pm \sqrt{2^2-4 \cdot (-(x-5))}}{2}$

$\displaystyle y = 1 \pm \sqrt{1+(x-5)}$

$\displaystyle y = 1 \pm \sqrt{x-4}$

Thanks! I would not have guessed that i needed to apply the Quadratic equation to find the inverse. I was under the impression that when finding inverses algebraically we interchange the x and y variables and then make y the subject.... Thanks again!