Question with limits and absolute max /min?

Question

I am working on a problem and would like to check my answer. I'd appreciate your help!

y = (x^2)/(x^2 - 4)

A. Find the limit of y as x approaches +/- infinity. I think it's 0?

B. What are the missing points in the domain? I think they are 2, -2, 0.

C. Where dy/dx is greater than 0, equal to 0, less than 0. Not sure I know what to do here...

D. What is the absolute max / min on the closed interval (0,2)? I don't understand the interval... I thought the min was 0 and max 2. I'm lost?

Thanks very much!!

Answer 1

To work out (a), you need to use what's known as L'Hopital's Rule, which you can apply to any limits that work out to be 0/0 or intinity/infinity. Basically, it states that, in these cases, the limit of f(x) / g(x) is equal to the limit of f'(x) / g'(x), or their derivatives:

y' = 2x / 2x = infinity/infinity
y'' = 2 / 2 = 1 = Your limit!

For (b), missing points in the domain would be where y is undefined, in this case, where y approaches +/- infinity. 0 is included in the domain and produces the point (0,0)

(c) dy/dx is "the derivative of y with respect to x." If you've worked with derivatives before, you should remember the Quotient Rule here ([lodihi - hidilo] / lolo; hi is the numerator, lo is the denominator, and di is a short way to remember the derivative). In this case,
dy/dx = y' = (2x(x^2 - 4) - x^2 (2x)) / (x^2 - 4)^2
dy/dx = -8x / (x^2 - 4)^2

To find (d), the maxima and minima on a given interval occur where the derivative of the function within that interval is equal to zero. For example, to find the minimum of x^2 on the closed interval of {-1, 1} calculate its derivative (2x) and solve for 0 (which occurs at x = 0).
To determine whether it's a maximum of minimum, form a number line and check whether the derivative is positive or negative on either side of the number. In this case, the derivative is negative for all x less than 0 and positive for all x greater than zero. This shows us that the point (0, 0) on the curve y = x^2 is a minimum.

Answer 2

A. The limit as x approaches +-infinity is 1.
B. Missing points are just 2 and negative 2.
C. Derivative = (-8X)/(x^2-4)^2: When x>0 slope is negative; when x<0, slope is positive; when x=0, slope is zero; when x=+-2, the slope is undefined.
D. Between x=0 and x=2, what is the greatest value, answer:0.

Answer 3

A) y=1 when x=infinity
When you do this, you can disregard any term without an x in it, since they will be infinitely small compared to the x terms, so you get x^2/x^2=1
B)The missing points are -2 and 2. 0 isn't missing it will just give y=0.