Functions question help me?

Question

a) We have the function $f(x,y)=x-y+1.$ Find the values of $f(x,y)$ in the points of the parabola $y=x^2$ and build the graph $F(x)=f(x,x^2)$ .

So, some points of the parabola are $(0;0), (1;1), (2;4)$. I replace these in $f(x,y)$ and I have $f(x,y)=1,1,-1\dots$. The graph $f(x,x^2)$ must have the points $(0;1),(1;1)$ and $(2;-1)\,$ , right?

b)We have the function $f(x,y) =\sqrt x + g[\sqrt x-1]$.Find $g(x)$ and $f(x,y)$ if $f(x,1)=x$?

I dont even know where to start here :/

Out of the subject,Thomas,I love programming in Java too haha! — none85, Mar 19 '13 at 17:39
I hope I didn't mangle your question while re-formatting. Please check. — Thomas Andrews, Mar 19 '13 at 17:39

score 3 · Answer 1 · answered Mar 19 '13 at 17:48

3

Hints: a) $F(x)=f(x,x^2)=x-x^2+1 =-(x-\frac12)^2+...$

b) As it stands, if I understand well, $f(x,y)=\sqrt x+g(\sqrt x-1)$ is independent from $y$. Then $f(x,1)=x$ means $$\sqrt x+g(\sqrt x-1)=x\,.$$ To find $g$, probably the best is to introduce a new variable $z:=\sqrt x-1$, then $\sqrt x=z+1$ and $x=(z+1)^2$.

answered Mar 19 '13 at 17:48

Berci

90,745

Isn't that $,g[\sqrt x-1] = g,$ of the integral value of $,\sqrt x-1,$ ? – DonAntonio Mar 19 '13 at 17:50
Possibly yes, but that would make the exercise harder.. – Berci Mar 20 '13 at 10:43

score 3 · Answer 2 · answered Mar 19 '13 at 17:51

3

And If the function $f(x,y)$ be like $$f(x,y)=\sqrt{y}+g(\sqrt{x}-1)$$, then the way is similar to @Berci's answer.Y ou pointed that $f(x,1)=x$ so $$1+g(\sqrt{x}-1)=x$$ Set $\sqrt{x}-1=t$ then $x=(t+1)^2$ and so $$g(t)=(t+1)^2-1=t^2+2t$$ Now you can easily find the function $f(x,y)$.

answered Mar 19 '13 at 17:51

Mikasa

67,374

Oh, Babak, I miss you tonight! I'm doing terribly tonight! Will have to go to bed disappointed with myself :-($\quad$ Perhaps tomorrow will be a fresh, new day for me! – amWhy Mar 20 '13 at 05:13
You can't be online at all, ever? Or now...or tomorrow? – amWhy Mar 20 '13 at 05:39
@amWhy: I can but just for seconds. Then I get disconnected. Again for some seconds later I come..... till tomorrow I think. It is OK Amy. ;-) – Mikasa Mar 20 '13 at 06:01
+1 For taking into account that it is highly suspicious to have a function of two variables without one of the variables playing a role... – DonAntonio Mar 20 '13 at 13:10

amWhy · Answer 3 · 2013-03-19T18:18:52.237

(a)

To find points on $f(x, y)$ that are also on the parabola $y = f(x) = x^2$: Solve for where $f(x, y) = x - y + 1$ and $y = f(x) = x^2$ intersect by putting $f(x, y) = f(x)$: $$ x^2 = x - y + 1$$ and and express as a value of $y$:

$$y = 1 + x - x^2\;\;\text{ and note}\;\; y = F(x) = f(x, x^2)\tag{1}$$

Note that this function $F(x)$ is precisely $f(x, x^2)$. Find enough points in this equation to plot it. (The points that satisfy $(1)$ will be points on $f(x,y)$ which satisfy $y = x^2$.) You will find that $\;\;F(x) = 1 + x - x^2\;\;$ is also a parabola. (See image below.) We can manipulate $F(x)$ to learn where the vertex of the parabola is located: write $F(x): (y - \frac 54) = -(x - \frac 12)^2$, so the vertex is located at $(\frac 12, \frac 54)$. The negative sign indicates that the parabola opens downward.

$\quad F(x) = 1 + x - x^2:$

enter image description here

Functions question help me?

3 Answers3