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December 23rd, 2015, 04:38 AM   #1
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Self-inverse function

A function is said to be a self-inverse if f(x)=f^-1(x) for all x in the domain.
a) Show that f(x)=1/x is a self-inverse function.
b) Find the value of the constant k so that g(x)=(3x-5)/(x+k) is a self-inverse function.

A) How should I prove that: y=1/x
yx=1
x=1/y
f^-1(x)=1/x
But in this way it does not make sense.

B) ??

Last edited by skipjack; December 23rd, 2015 at 04:44 AM.
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December 23rd, 2015, 04:43 AM   #2
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(a) If y = 1/x, x = 1/y, so the inverse of f(x) = 1/x is f$\,^{-1}$(x) = 1/x.

If you don't understand that, what exactly is puzzling you?
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Last edited by skipjack; December 23rd, 2015 at 04:46 AM.
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December 23rd, 2015, 04:55 AM   #3
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I just wanted to check for the first one, because it seemed extremely easy, but it carries the highest amount of points. I thought I was wrong but now it seems I am right.

The second problem. How should I find the k?

Last edited by skipjack; December 23rd, 2015 at 05:15 AM.
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December 23rd, 2015, 05:20 AM   #4
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(b) Solving x = (3g - 5)/(g + k) for g gives g = (-kx - 5)/(x - 3), so k = -3 for g to be self-inverse.
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December 23rd, 2015, 05:41 AM   #5
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Formally, you find solution for $k$ that are independent of $x$ to
$$\begin{aligned}
{3x+5 \over x+k} &= {-kx+5 \over x-3} \\
(3x+5)(x-3) &= -(kx-5)(x+k) \\
3x^2 -4x -15 &= -kx^2 -(k^2 -5)x +5k \\
(3+k)x^2+(k^2-9)x-(15+5k) &= 0
\end{aligned}$$
The only way this can be true for all $x$ (independent of $x$) is if all the coefficients are zero.
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December 23rd, 2015, 06:24 AM   #6
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I understand thank you

Last edited by med1student; December 23rd, 2015 at 06:33 AM.
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December 23rd, 2015, 06:46 AM   #7
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It should be -kx-5,not -kx+5
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December 23rd, 2015, 06:49 AM   #8
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The crucial point of "inverse functions" is that $\displaystyle f(f^{-1}(x))= x$ and $\displaystyle f^{-1}(f(x))= x$. So a function, f, is "self inverse" if and only if $\displaystyle f(f(x))= x$.

If $\displaystyle f(x)=\frac{1}{x}$ then $\displaystyle f(f(x))= \frac{1}{\frac{1}{x}}= \left(1\right)\left(\frac{x}{1}\right)= x$ and we are done.

$\displaystyle g(x)= \frac{3x- 5}{x+ k}$ then g is "self inverse" if and only if $\displaystyle g(g(x))= \frac{3\left(\frac{3x- 5}{x+ k}\right)- 5}{\frac{3x- 5}{x+ k}+ k}= \frac{\frac{9x- 15}{x+ k}- 5}{\frac{3x- 5}{x+ k}+ k}$

Now multiply both numerator and denominator by x+ k to get
$\displaystyle g(g(x))= \frac{(9x- 15)- (5x+ 5k}{(3x- 5)+ kx+ k^2}= \frac{4x- 15- 5k}{(3+ k)x+ k^2- 5}$. That will be equal to x if and only if $\displaystyle 4x- 15- 5k= (3+ k)x^2+ (k^2- 5)x$. Since the left side has no "$\displaystyle x^2$", in order for these to be equal for all x, the right side must not either- we must have 3+ k= 0 or k= -3. In that case, 15- 5k= 15- 15= 0 also so we have 4x= (9- 5)x= 4x.
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December 23rd, 2015, 03:21 PM   #9
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Follow-up question: if f(x) = (ax + b)/(cx - a) is a self-inverse function, what condition must be satisfied by the coefficients a, b and c?
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