Finding x/y-intercepts, asymptotes, derivatives, and max min

In summary: I think. So I'm looking at this like, 64x^2 is multiplied by (x^2-4) and 64x^2 is multiplied by (x^2-4)^2, so 64x^2 is multiplied by (x^2-4) twice so it becomes 64x^2(x^2-4)^2. And this answer is inside the bracket with -16(x^2-4)^2, so it becomes -16(x^2-4)^2+64x^2(x^2-4)^2, which is the numerator. Right?Yes, exactly. This is the distributive law.
  • #1
Specter
120
8

Homework Statement



For the following function ##f(x)=\frac {2x^2} {x^2-4}##, find the following:

a) The x and y intercepts
b) the horizontal and vertical asymptotes
c) the first and second derivatives
d) any local maximum or minimum points
e) the intervals of increasing and decreasing
f) any inflection points
g) the intervals of concavity

(solved some of these already but included the full questions just incase)

Homework Equations

The Attempt at a Solution


[/B]
I mostly need help with the second derivative.

a) The x and y intercepts are (0,0)

b) The horizontal and vertical asymptotes are

Vertical:

##x^2-4=0##
##(x+2)(x-2)##
x=2 and x=-2

Horizontal:
##\frac 2 {1}=2##

c) The first and second derivatives

I had a lot of trouble finding the second derivative, and still don't understand a few of the final steps. Here's my work.

##f(x)=\frac {2x^2} {x^2-4}##

##f'(x)=\frac {4x(x^2-4)-2x^2(2x)} {(x^2-4)^2 }##

##=\frac {4x^3-16x-4x^3} {(x^2-4)^2}##

##=\frac {-16} {(x^2-4)^2}##

##f''(x)=\frac {-16(x^2-4)^2-(-16x)(4x(x^2-4))} {(x^2-4)^4}## I solved up to here on my own, after this I am not sure where ##(x^2-4)## went, and why the denominator went from ##(x^2-4)^4## to ##(x^2-4)^3##. Also where did the square go on the ##(x^2-4)^2##?

##=\frac {-16(x^2-4)+64x^2} {(x^2-4)^3}##

##=\frac {-16x^2+64+64x^2} {(x^2-4)^3}##

##=\frac {48x^2+64} {(x^2-4)^3}##

d) any local maximum or minimum points.

Set the first derivative to 0 solve for x:

##f'(x)=\frac {-16x} {(x^2-4)^2}=0##

##=-16x=0##

##x=0##

Sub x value into original function

##y=\frac {2(0)^2} {(0)^2-4}##

##=\frac 0 {-4} =0##

(0,0) is a possible max or min point.
 
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  • #2
Simple matter of factoring.

You have ##-16(x^2 - 4)^2 - (-16x)(4x)(x^2 - 4) = -16(x^2 - 4)^2 + 64x^2(x^2 - 4)##. Both terms have a ##(x^2 - 4)## factor, so you can factor it out.
## = (x^2 - 4) \left [ -16(x^2 - 4) + 64x^2 \right ]##
And that ##(x^2 - 4)## cancels out with one factor in the denominator, leaving you ##(x^2 - 4)^3## in the denominator and ##\left [ -16(x^2 - 4) + 64x^2 \right ]## in the numerator.

I think that answers the question you were asking.

If it's not clear, replace the expression ##x^2 - 4## with ##y##.
##\frac {-16y^2 + 64x^2y} {y^4} = \frac{y (-16y + 64x^2)} {y^4} = \frac{-16y + 64x^2}{y^3}##
 
  • #3
RPinPA said:
Simple matter of factoring.

You have ##-16(x^2 - 4)^2 - (-16x)(4x)(x^2 - 4) = -16(x^2 - 4)^2 + 64x^2(x^2 - 4)##. Both terms have a ##(x^2 - 4)## factor, so you can factor it out.
## = (x^2 - 4) \left [ -16(x^2 - 4) + 64x^2 \right ]##
And that ##(x^2 - 4)## cancels out with one factor in the denominator, leaving you ##(x^2 - 4)^3## in the denominator and ##\left [ -16(x^2 - 4) + 64x^2 \right ]## in the numerator.

I think that answers the question you were asking.

If it's not clear, replace the expression ##x^2 - 4## with ##y##.
##\frac {-16y^2 + 64x^2y} {y^4} = \frac{y (-16y + 64x^2)} {y^4} = \frac{-16y + 64x^2}{y^3}##

That kind of makes sense. What confuses me is the square on ##(x^2-4)^2##. Now this is my first math course in a while, but wouldn't it just get factored from ##(x^2-4)^2## to##(x^2-4)## or is that not how it works?

This is what I was thinking, but I haven't had to factor in a while so I'm probably wrong.

##f''(x)=\frac {-16(x^2-4)^2+64x^2(x^2-4)} {x^2-4)^4}##

##=\frac {(x^2-4) [-16(x^2-4)^2+64x^2} {(x^2-4)^4}##

##=\frac {-16(x^2-4)+64x^2} {(x^2-4)^3}##
 
  • #4
Go the other way. What is ##y (-16y^2 + 64x^2)##? By the distributive law it's ##-16y^2*y + 64x^2*y = -16y^3 + 64x^2y##

Do that with your expression. Multiply the ##x^2 - 4## back by both of the terms inside the square brackets. What happens to the first term when you multiply it by ##(x^2 - 4)##? What is ##-16(x^2 - 4)^2## times ##(x^2 - 4)##?

I think I'm not quite understanding your question. You asked "wouldn't it get factored from ##(x^2 - 4)^2## to ##(x^2 - 4)##? If you look at what I did, I had ##(x^2 - 4)^2## before factoring and ##(x^2 - 4)## inside the brackets after factoring. So it did in fact get factored from ##(x^2 - 4)^2## to ##(x^2 - 4)##.
 
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  • #5
RPinPA said:
Go the other way. What is ##y (-16y^2 + 64x^2)##? By the distributive law it's ##-16y^2*y + 64x^2*y = -16y^3 + 64x^2y##

Do that with your expression. Multiply the ##x^2 - 4## back by both of the terms inside the square brackets. What happens to the first term when you multiply it by ##(x^2 - 4)##? What is ##-16(x^2 - 4)^2## times ##(x^2 - 4)##?

I think I'm not quite understanding your question. You asked "wouldn't it get factored from ##(x^2 - 4)^2## to ##(x^2 - 4)##? If you look at what I did, I had ##(x^2 - 4)^2## before factoring and ##(x^2 - 4)## inside the brackets after factoring. So it did in fact get factored from ##(x^2 - 4)^2## to ##(x^2 - 4)##.
Thank you, it just took some time for me to understand this.I get it now!
 

Related to Finding x/y-intercepts, asymptotes, derivatives, and max min

1. What is the difference between x and y-intercepts?

The x-intercept is the point at which a graph crosses the x-axis, meaning that the value of y is equal to 0. The y-intercept is the point at which a graph crosses the y-axis, meaning that the value of x is equal to 0.

2. How do you find the x and y-intercepts of a function?

To find the x-intercept, set y=0 in the equation and solve for x. To find the y-intercept, set x=0 in the equation and solve for y.

3. What is an asymptote and how do you find it?

An asymptote is a line that a graph approaches but never touches. It can be vertical, horizontal, or oblique. To find an asymptote, set the denominator of the function equal to 0 and solve for x. The resulting value of x will give the equation of the vertical asymptote. For horizontal asymptotes, take the limit as x approaches infinity or negative infinity. If the limit exists and is a finite number, then that value is the equation of the horizontal asymptote.

4. How do you find the derivative of a function?

The derivative of a function represents the rate of change of the function at a specific point. To find the derivative, use the power rule for functions with exponents, the product rule for functions that are multiplied, and the quotient rule for functions that are divided. For functions that are more complex, you may need to use the chain rule or other differentiation rules.

5. What are the maxima and minima of a function and how do you find them?

The maxima and minima of a function represent the highest and lowest points on the graph, respectively. To find them, take the derivative of the function and set it equal to 0. Solve for x to find the x-coordinate of the maxima or minima. To determine if it is a maxima or minima, you can use the second derivative test or plug in values of x on either side of the critical point to see if the function is increasing or decreasing.

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