Inequalities

**arze** · September 8th 2010, 08:30 PM

The equation of a curve is $x^2y^2-x^2+y^2=0$
a) find the equation of the tangents at the origin
b) find the equations of the real asymptotes
c) show that the numerical value of y is never greater than the corresponding value of x
d)show that the numerical value of y is always less than unity

My problem is with (c) and (d). I've done (a) and (b) already.
For (c), $x-y\geq 0$
$y=\pm\frac{x}{\sqrt{x^2+1}}$
Substitute
$x-\frac{x}{\sqrt{x^2+1}}=\frac{x\sqrt{x^2+1}-x}{\sqrt{x^2+1}}$
Now I don't know how to continue to make it such that it is greater or equal to 0, which then i can show that y is always lesser of equal to x.
Same problem with the next one. The $\sqrt{x^2+1}$ is what stumble me.
Thanks!

**Prove It** · September 9th 2010, 12:10 AM

I think you should rewrite this as

$y = \left|\frac{x}{\sqrt{x^2 + 1}}\right|$ .

Then to show that $y \leq x$ , you would have

$\left|\frac{x}{\sqrt{x^2 + 1}}\right| \leq x$

$-x \leq \frac{x}{\sqrt{x^2 + 1}} \leq x$

$-1 \leq \frac{1}{\sqrt{x^2 + 1}}\leq 1$ .

This would only ever not be true if the denominator was smaller than the numerator. So it would not be true if $\sqrt{x^2 + 1} < 1$

If $x^2 + 1 < 1$ then

$x^2 < 0$ .

Since this is never true, that means that

$-1 \leq \frac{1}{\sqrt{x^2 + 1}} \leq 1$ IS true.

Therefore $y \leq x$ for all $x$ .

To answer d) you want to show that

$y = \left|\frac{x}{\sqrt{x^2 + 1}}\right| < 1$ .

Therefore $-1 < \frac{x}{\sqrt{x^2 + 1}} < 1$

$-\sqrt{x^2 + 1} < x < \sqrt{x^2 + 1}$

This should be obvious, since $x = \sqrt{x^2}$ .

$\sqrt{x^2} < \sqrt{x^2 + 1}$ .

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