# Finding the range of a rational function

1. Jan 24, 2015

### sooyong94

1. The problem statement, all variables and given/known data
A curve is given by the parametric equations
$x=t^2 +3$
$y=t(t^2+3)$
Find dy/dx in terms of t and show that (dy/dx)^2 >=9

2. Relevant equations
Parametric derivatives

3. The attempt at a solution
Using the chain rule, I arrived at $\frac{dy}{dx}=\frac{3}{2}(\frac{t^2+1}{t})$
However, when I squared both sides to get $(\frac{dy}{dx})^2$, I was unable to prove that (dy/dx)^2>=9. I know I need to find the range, however I'll need to graph the function, which proves to be tedious. Is there any workaround to this?

2. Jan 24, 2015

### PeroK

What did you try? Isn't it just some simple quadratic algebra?

3. Jan 24, 2015

### sooyong94

I was thinking to graph the function, though I'll be dealing with quartic functions.

4. Jan 24, 2015

### PeroK

Some quartics look like quadratics, wouldn't you say?

5. Jan 24, 2015

### sooyong94

Yup, but how do I begin though?

6. Jan 24, 2015

### PeroK

If $(dy/dx)^2 < 9$ what does that give you?

7. Jan 24, 2015

### sooyong94

There are no values of t that satisfies the inequality?

8. Jan 25, 2015

### PeroK

Yes, but you've got to be prepared to do the algebra that shows why.

9. Jan 25, 2015

### sooyong94

So how do I begin?

10. Jan 25, 2015

### Staff: Mentor

Your problem is equivalent to showing that $(\frac{t^2 + 1}{t})^2 \ge 4$.
Presumably you are in a calculus class. How do you find the minimum value of a function?

Last edited: Jan 25, 2015
11. Jan 25, 2015

### sooyong94

Derive and set the value of the derivative to 0?

12. Jan 25, 2015

### Staff: Mentor

Yes, differentiate it and set the derivative to 0. But how will you know whether you have a minimum or maximum?

Now show us something...

13. Jan 25, 2015

### sooyong94

But first, why did you say ((t^2+1)/t)^2 >4?

Use the first derivative test for critical points?

14. Jan 25, 2015

### Staff: Mentor

I changed what I wrote slightly, to $(\frac{t^2 + 1}{t})^2 \ge 4$
What do you think? Take a stab at it.

15. Jan 25, 2015

### sooyong94

But what makes you think why ((t^2 +1)/t))^2 >=4?

The sign of the slopes changes?

16. Jan 25, 2015

### Staff: Mentor

In your first post you have
Isn't the inequality on the right equivalent to the one I wrote?

17. Jan 25, 2015

### sooyong94

Wasn't it should be >=9?

18. Jan 26, 2015

### Staff: Mentor

You tell me.
2. Simplify.

This is post 18 in the thread. So far I haven't seen any work from you. What's stopping you?

19. Jan 26, 2015

### sooyong94

I have two inequalities, 3/2((t^2 +1)/t) <=-3 or ((t^2 +1)/t) >=2. Then how should I proceed from here?

20. Jan 26, 2015

### Staff: Mentor

$(dy/dx)^2 \ge 9$
$\Rightarrow (3/2)^2(\frac{t^2 + 1}{t})^2 \ge 9$
$\Rightarrow (\frac{t^2 + 1}{t})^2 \ge 4$
This last inequality is equivalent to $\frac{t^2 + 1}{t} \ge 2$ or $\frac{t^2 + 1}{t} \le -2$
Since this is what you need to show, you can't assume that it's true. You can use calculus techniques to find the minimum and values of the rational expression.