Is this is the right way to do these one-to-one functions, finding their inverse, if not, how to do it?

Question

Question 1) $f(x) = 1-x$

My answer (1): $f(x) = 1-x$, $y = 1 - x$, $y + 1 = x$, $x = y + 1$, $f$ of inverse $f(y) = y + 1$

Question 2) : $f(x) = \dfrac{2x}{x-1}$

My answer 2) : $f(x) = \frac{2x}{x-1} = y$, $\frac{2x}{x-1}$, $y+1=2x$, $\frac{y+1}{2} = x$, $x= \frac{y+1}{2}$, $f$ of inverse $f(y) = \frac{y+1}{2}$

Question 3): $f(x) = \sqrt{5} - x$

My answer 3) : $f(x) = \sqrt{5} - x = y = \sqrt{5} - x = y + 5 = x = x = y + 5$, f of inverse $f(y) = y + 5$

Question 4): $f(x) = x^3$

My answer 4): $f(x) = x^3$, $y=x^3$, $y^3 = x$, $x=y^3$, f of inverse $f(y) = y^3$

Question 2: do you mean $\frac{2x}{x-1}$? Question 3: do you mean $\sqrt{5}-x$? — wythagoras, May 16 '15 at 10:31
Please see this tutorial on how to typeset mathematics on this site. Also, I suggest you lay out your work vertically. If you read your answer to 3), you will notice that you obtained $y = y + 5$. — N. F. Taussig, May 16 '15 at 11:44

wythagoras · Accepted Answer · 2015-05-16T11:20:49.270

2

Question 1. I believe the method is right, but there is a small mistake.

\begin{align*} y &= 1-x \\ y-1 &= -x \\ 1-y &= x \end{align*}

Question 2.

\begin{align*} y &= \frac{2x}{x+1} \\ \frac{y}{2} &= \frac{x}{x+1} \\ \frac{y}{2} &= 1-\frac{1}{x+1} \\ \frac{y}{2}-1 &= -\frac{1}{x+1} \\ 1-\frac{y}{2} &= \frac{1}{x+1} \\ \frac{1}{1-\frac{y}{2}} &= x+1 \\ \frac{1}{1-\frac{y}{2}}-1 &= x \\ \end{align*}

Question 3. This is similiar to question 1, can you do this yourself?

\begin{align*} y &= \sqrt{5}-x \\ y-\sqrt{5} &= -x \\ \sqrt{5}-y &= x \end{align*}

Question 4.

\begin{align*} y &= x^3 \\ \sqrt[3]{y} &= x \end{align*}

edited May 16 '15 at 11:20

answered May 16 '15 at 10:29

wythagoras

25,026

hmmm, thanks, how about number 2 and 3? – Abdiaziz Abdullahi May 16 '15 at 10:36
@HeeganHero See my comment under your question. – wythagoras May 16 '15 at 10:44
I have done 2 since that one is quite hard. – wythagoras May 16 '15 at 10:51
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yes, I can do number 3, thank you very much :) – Abdiaziz Abdullahi May 16 '15 at 10:54
You're welcome. By the way, if you like the answer, you can also upvote it, in addition to accepting it. – wythagoras May 16 '15 at 10:56
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;( VoteUp requires 15 reputation, getting that error, but I'll do it after reaching that point :) – Abdiaziz Abdullahi May 16 '15 at 11:03
@ wythagoras , I tried number 3, is this right please ,f(x) = √5−x , y = √5−x = -x , 5-y = x , x= 5-y – Abdiaziz Abdullahi May 16 '15 at 11:10
Sorry, I always forget about the reputation thresholds. Regarding question 3, why is the root suddenly gone? – wythagoras May 16 '15 at 11:20
Can you do f(x)=3-x? – wythagoras May 16 '15 at 11:21
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hhh you're right, I supposed not to take that shortcut, on the other way, F(x)=3-x, y=3-x=-x , 3-y=x, x=3-y – Abdiaziz Abdullahi May 16 '15 at 11:31
You found the correct function, but please, do not write y=3-x=-x, because that implies 3-x=-x, which is not true. – wythagoras May 16 '15 at 11:42
hmmm, now I know that, you are very helpful, thanks indeed. – Abdiaziz Abdullahi May 16 '15 at 11:45

Is this is the right way to do these one-to-one functions, finding their inverse, if not, how to do it?

1 Answers1