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

[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

 

[danago]

I don't 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?

 

[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!

 

[HallsofIvy]

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


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

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

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!

 

[statdad]

Your line

<br /> y = \frac 1 {3+x^2} = 3+x^{-2}<br />

is the source of your error.

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

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

 

[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!