# Curve Sketching, Concavity and pts of Inflection

1. Aug 25, 2008

### roam

Let $$m(x) = \frac{1-x^2}{x^3}$$

Sketch the graph and find all critical points and any points of inflection.

3. The attempt at a solution
$$m(x) = \frac{1-x^2}{x^3}$$

$$\frac{(x^3 . -2x) - (1-x^2) . (3x^2)}{(x^3)^2}$$ (quotient rule)

using the power rule;

$$\frac{-2x^4 - (3x^2 + 3x^4)}{x^6}$$

$$\frac{x^4 - 3x^2}{x^6}$$

$$\frac{x^2(x^2 - 3)}{x^6}$$

$$m'(x) = \frac{x^2 -3}{x^4}$$

Therefore the critical numbers must be: 0, $$\pm \sqrt{3}$$

Is it correct so far?

Now to find the points of inflection is where I'm stuck atm;

We apply the quotient rule again;

$$m''(x) = \frac{(x^4 . 2x) - ((x^2 -3) . 4x^3)}{(x^4)^2}$$

$$m''(x) = \frac{(x^4 . 2x) - ((x^2 -3) . 4x^3)}{(x^4)^2} = 0$$

How can I simplify this so it's easier to obtain the possible points of inflection?

P.S. I figured that this function has a horizontal asymptote, y= 0. But when I draw it on calculator or matlab the function crosses this line. Why?

2. Aug 25, 2008

### Hitman2-2

Your calculation of the first derivative and the critical points look right to me.

To simply the second derivative, just do what you did in finding the first derivative: multiply out the terms in the numerator, then simplify.

As to why the graph crosses the x-axis, a function can actually cross the asymptote as long as it eventually approaches the asymptote arbitrarily closely and stays close (informally speaking).

Additionally, if I may make a suggestion, when graphing functions, it can be helpful to first find the domain, whether the function is even or odd, and x and y intercepts. Here, for example, the function is odd, so you only need to be concerned about non-negative x (at first) and then sketch the graph for negative x by symmetry.

Last edited by a moderator: Aug 25, 2008
3. Aug 25, 2008

### HallsofIvy

Staff Emeritus

Why haven't you multiplied it out just like you did the first derivative?
$$m''(x)= \frac{2x^5- 4x^5+ 12x^3}{x^8}= \frac{-2x^5+ 12x^3}{x^8}= \frac{-2x^2+ 12}{x^5}$$

What's the problem? There is no reason a rational function can't cross a horizontal asymptote. A horizontal asymptote only says what happens as x goes to plus or minus infinity.

4. Aug 26, 2008

### roam

Yea right.

$$m''(x) = \frac{-2x^2+ 12}{x^5}$$

$$\Rightarrow \frac{-2(x-3)^2}{x^5}$$

Therefore the x coordinates of the inflection points would be 0, 3 ? So I have to insert these x values into g'(x) to find their y coordinates?

I'm sure that there are two points of inflection because when I draw the graph there are two points at wich the function changes concavity.

5. Aug 26, 2008

### Hitman2-2

Are you saying that

$$m''(x) = \frac{-2x^2+ 12}{x^5} = \frac{-2(x-3)^2}{x^5} ?$$

The second equality above isn't true, so 3 is definitely not a point of inflection.

6. Aug 26, 2008

### HallsofIvy

Staff Emeritus
No, that's not at all right.
$$\frac{-2x^2+ 12}{x^5}= -2\frac{x^2- 6}{x^5}[/itex] There are 2 points where the second derivative is 0. Whether they are points of inflection or not depends on whether the second derivative changes sign there. By the way, x= 0 gives neither a critical point nor a point of inflection since the original function is not defined there. 7. Aug 26, 2008 ### roam Oops! I think those two points are [tex]\pm\sqrt{6}$$ because that's the only place m'' = 0

& I understand that x=0 is not a critical pt etc since the original function isn't defined there, like you said.

Thanks.

So, all I need to do now is to plug these x-coordinates into the original m(x) to find their y coordinates?

Thanks Hitman. Bleh, I knew that sorry I made a mistake!

Last edited: Aug 26, 2008
8. Aug 26, 2008

### Hitman2-2

Yep; you should be good to go.

9. Aug 28, 2008

### roam

We have inflection points at x = sqrt(6) and x = -sqrt(6).
to find the y-coordinate as well;

f(sqrt(6)) = (1 - 6) / (6 sqrt(6) )
= (-5/6) (1/sqrt(6))
= (-5/6) (sqrt(6)/6)
= -5sqrt(6)/36

f(-sqrt(6)) = (1 - 6) / ( -6sqrt(6))
= 5sqrt(6)/36

So the inflection points are at $$(\sqrt{6}, \frac{-5\sqrt{6}}{36})$$ and $$(-\sqrt{6}, \frac{5\sqrt{6}}{36})$$.

When I draw the graph I understand for which reigons the graph is concave up or concave down. Can anyone show me how to calculate it as well? ...because it does say find the intervals on which m is concave up/down...

10. Aug 28, 2008

### roam

Is this valid;

drawing a number line consisting of all critical values.

. . . . . . . . -sqrt(6) . . . . . . . .{0} . . . . . . . . . sqrt(6) . . . . . . . . . . .

By testing different values in those reigons I found that;

. . .{+}. . . . -sqrt(6) . . {-}. . . .{0} . . . . {+}. . . sqrt(6) . . . .{-}. . . . . .

I marked the positive and negative reigons. In the negative regions, our function is concave down; in the positive regions, concave up.

Am I right? ... I don't know how to write this anyway...

m(x) is concave up on $$(\infty, -\sqrt{6}) \bigcup (0, \sqrt{6})$$

m(x) is concave down on $$(-\sqrt{6}, 0) \bigcup (\sqrt{6}, \infty)$$

I don't know if it's right.

11. Aug 28, 2008

### rootX

looks good
http://img168.imageshack.us/img168/7426/ggih0.png [Broken]

Last edited by a moderator: May 3, 2017