# Graph

• Nov 1st 2009, 11:40 AM
Aquafina
Graph
I need hints on the following questions:

1. Sketch the curve (y2-2)2+(x2-2)2=2. What does it look like?

I got 4 circles which kind of merge together where they meet.

2. kx^4=x^3-x Find the real roots when k=0. Sketch the graph when k is small and then when k is large, and find approximations of the real roots in both cases. When else does x have 3 real roots?

I don't understand why the graph has 3 roots, even when k is not 0?

3. sketch x^x

The graph is a bit complicated, and broken down in the negative x values. I'm not sure how to draw it when x = 0, on the calculator it doesn't show it as an asymptote or anything, is it just supposed to end there or what? Also, I dont understand how to work out the fluctuations between positive and negative values of y for the negative values of x.

4. Sketch f(x) = (x - R(x))2, where R(x) is x rounded up or down in the usual way. then sketch g(x) = f(1/x)

For f(x) I got like multiple humps between the integer, with maximum at half way between 2 integers of 1/4, and then 0 at the integers. However, I have no idea for g(x), it doesn't seem to follow any pattern.
• Nov 3rd 2009, 06:42 AM
Hello Aquafina
Quote:

Originally Posted by Aquafina
I need hints on the following questions:

1. Sketch the curve (y2-2)2+(x2-2)2=2. What does it look like?

I got 4 circles which kind of merge together where they meet.

Note that we have just even powers of $x$ and $y$, so the graph is symmetrical about both axes. If we re-arrange:

$y^2=2\pm\sqrt{2-(x^2-2)^2}$

Note that whenever $\sqrt{2-(x^2-2)^2}$ is real, $2-\sqrt{2-(x^2-2)^2} \ge 0$ so $y$ is real.

Then $\sqrt{2-(x^2-2)^2}$ is real whenever $(x^2-2)^2\le 2$

i.e. $\sqrt{2-\sqrt2}\le x \le \sqrt{2+\sqrt2}$ or $-\sqrt{2+\sqrt2}\le x \le -\sqrt{2-\sqrt2}$.

These give approximate values of $0.76 \le x \le 1.85$ and $-1.85\le x \le - 0.76$.

By the symmetry of the equation, $y$ has the same ranges of values.

So the graph consists of four non-overlapping loops as in the attached sketch.