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

In summary, the conversation discusses finding the sketch for the function y = \frac{1}{3+x^2} and the work needed to find x and y intercepts, asymptotes, critical points and points of inflection. There is a discussion on finding the derivatives of the function and correcting an error in the first derivative. The final answer includes the correct derivatives and the points of inflection.
  • #1
lamerali
62
0
Sketch the following function, showing all work needed to sketch each curve.

y = [tex]\frac{1}{3 + x^2}[/tex]

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 [tex]^{-3}[/tex]

= [tex]\frac{-2}{x^-3}[/tex]

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

to find the point of infliction find the second derivative

y11 = 6x [tex]^{-4}[/tex]
= [tex]\frac{6}{x^-4}[/tex]

y11 can never equal zero
for x [tex]\geq[/tex] 0, y11 is negative and for x [tex]\leq[/tex] 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
 
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  • #2
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?
 
  • #3
That is exactly what confused me!
I found the derivatives as follows:


y = [tex]\frac{1}{3+x^2}[/tex]
= 3 + x[tex]^{-2}[/tex]

first derivative
y1 = -2x[tex]^{-3}[/tex]
= [tex]\frac{-2}{x^3}[/tex]

second derivative
y11 = -2x[tex]^{-3}[/tex]
= (-2)(-3)x [tex]^{-4}[/tex]
= 6x [tex]^{-4}[/tex]
= [tex]\frac{6}{x^4}[/tex]

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


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

first derivative
y1 = -2x[tex]^{-3}[/tex]
= [tex]\frac{-2}{x^3}[/tex]

second derivative
y11 = -2x[tex]^{-3}[/tex]
= (-2)(-3)x [tex]^{-4}[/tex]
= 6x [tex]^{-4}[/tex]
= [tex]\frac{6}{x^4}[/tex]

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!
 
  • #5
Your line

[tex]
y = \frac 1 {3+x^2} = 3+x^{-2}
[/tex]

is the source of your error.

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

which is obviously false. Try writing the formula for [tex] y [/tex] with the entire denominator having a negative exponent.
 
  • #6
YAY! thanks a lot! i think i got it!
for the critical points
the first derivative would be

y1 = [tex]\frac{-2x}{(3+x^2)^2}[/tex]

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 = [tex]\frac{6x^2 - 6}{(3 + x^2)^3}[/tex]

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!
 

What are the intercepts of a curve?

Intercepts are the points where a curve crosses the x-axis or the y-axis. To find the x-intercept, set y=0 and solve for x. To find the y-intercept, set x=0 and solve for y.

What are asymptotes?

Asymptotes are imaginary lines that a curve approaches but never touches. They can be vertical, horizontal, or oblique. Vertical asymptotes occur when the denominator of a rational function is equal to 0. Horizontal asymptotes occur when the degree of the numerator is less than the degree of the denominator.

What are critical points?

Critical points are the points where the derivative of a curve is equal to 0 or does not exist. They can be local maximums or minimums, points of inflection, or points of discontinuity.

How do you find the critical points of a curve?

To find the critical points, take the derivative of the curve and set it equal to 0. Solve for x to find the x-coordinates of the critical points. Then, plug these values into the original equation to find the y-coordinates.

How do you sketch a curve using intercepts, asymptotes, and critical points?

To sketch a curve, first plot the intercepts and any known points. Then, determine the behavior of the curve near the asymptotes. Finally, use the critical points to determine the overall shape of the curve and sketch it accordingly.

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