# Homework Help: Sketching curves: intercepts, asymptotes, critical points [answer check]

1. Sep 19, 2008

### lamerali

Sketch the following function, showing all work needed to sketch each curve.

y = $$\frac{1}{3 + x^2}$$

The question is asking for all the work done to find x and y intercepts, vertical, horizontal and slant asymptotes; critical points and points of inflection, i have completed the question but i'm not sure i did it correctly, any guidance is appreciated. My answer is below:

x - intercept: there are no x - intercepts.
y - intercept: sub x = 0 into the function and you get y = 1/3

there are no vertical asymptotes

horizontal asymptote is y = 0 found by dividing every term in the function by x^2

critical points:
to find the critical points find the first derivative

y1 = -2x $$^{-3}$$

= $$\frac{-2}{x^-3}$$

y1 can never equal zero therefore there are no max or min.

to find the point of infliction find the second derivative

y11 = 6x $$^{-4}$$
= $$\frac{6}{x^-4}$$

y11 can never equal zero
for x $$\geq$$ 0, y11 is negative and for x $$\leq$$ 0, y11 is positive

I have sketched the graph but not added it here i just would like to check if the work i did to get the sketch of the function is correct!
Thanks A LOT!
I really appreciate it!! :D

2. Sep 19, 2008

### danago

I dont understand how you found the derivative of y. Check that again, because if you graph the function, you see that there is indeed a maximum at x=0.

EDIT: Same with your second derivative. How exactly are you differentiating these?

3. Sep 19, 2008

### lamerali

That is exactly what confused me!
I found the derivatives as follows:

y = $$\frac{1}{3+x^2}$$
= 3 + x$$^{-2}$$

first derivative
y1 = -2x$$^{-3}$$
= $$\frac{-2}{x^3}$$

second derivative
y11 = -2x$$^{-3}$$
= (-2)(-3)x $$^{-4}$$
= 6x $$^{-4}$$
= $$\frac{6}{x^4}$$

for both of these y cannot equal zero...therefore there are no critical points or points of infliction. however, i too see that there is a max at about (0, 1/3)!!
any idea of where i am going wrong!
thanks!

4. Sep 19, 2008

### HallsofIvy

No. $$y= (3+ x^2)^{-1}$$ You will need to use the chain rule to find the derivative.

5. Sep 19, 2008

$$y = \frac 1 {3+x^2} = 3+x^{-2}$$

is the source of your error.

Look at what happens for $$x = 1$$ with your approach
$$\frac 1 {3 + 1^1} = 3 + 1^{-1} = 4$$

which is obviously false. Try writing the formula for $$y$$ with the entire denominator having a negative exponent.

6. Sep 20, 2008

### lamerali

YAY! thanks a lot! i think i got it!
for the critical points
the first derivative would be

y1 = $$\frac{-2x}{(3+x^2)^2}$$

which is equal to zero when x = 0 forming the point (0, 1/3) which i found to be a max.

for the second derivative and the point of inflection
y11 = $$\frac{6x^2 - 6}{(3 + x^2)^3}$$

y11 is equal to zero when x = 1 and x = -1. I found the points of inflection to be (1, 1/4) and (-1, 1/4).

I'm pretty sure this is right i'd just like to make sure!
thank you!