Attached. Please just reply if you have any questions.
Correct sketch: upper left.
a. x intercept: set y = 0 and solve for x
0 = x^2 + x^-2
y intercept: set x = 0 and solve for y.
y = 0^2 + 1/0^2
x int (x,y) = DNE
y int (x,y) = DNE
b. To find extrema, take the derivative, set equal to zero to find the critical points.
y ' = 2x− 2x
0= 2x− 3
x= ± 1
You can then use the first derivative test to determine what type of extrema this is.
Pick points on either side to determine if the function is increasing or decreasing. Remember to
also divide the interval at vertical asymptotes (x = 0)
(-inf, -1) f ' (-2) = -3.75 DECREASING
(-1, 0) f ' ( -0.5) = 15
(0, 1) f ' (0.5) = -15
(1, inf) f ' (2) = 3.75
At x = -1, the function goes from decreasing to increasing making that a local minimum
At x = 1, the function goes from decreasing to increasing making that a local minimum
relative minimum (x,y) = (-1,2)
relative minimum (x,y) = (1,2)
c. To find possible inflection points, find the second derivative and set that equal to zero.
y ' ' = 2+ 6x
0= 2+ 4
− 2= 4
− 2x = 6
x =− 3
Since you cannot take the even root of a negative number, there will be no solution. Therefore
there can be no inflection points.
(x,y) = DNE
d. First determine where the original function is undefined. This will be where the
denominator is equal to zero. That point is when x = 0.
Then determine how the function behaves as you get closer and closer to zero. You can do this
by plugging a few values in for x.
y(0.5) = 4.25
y(0.1) = 100.01
y(0.01) = 10000.0001
You can see from that, as x gets closer to zero, the function is getting much larger.
We could have also looked at the first derivative from part b. In the 2 intervals approaching 0
from the left and right. From the left the function is increasing, going toward infinity. From the
right going from x = 0 to x = 1, the function is decreasing that means that moving back toward 0
it would be increasing.
Y → + infinity, as x →...