For the following function, find the
Domain: all reals but 0
Symmetry: Origin Symmetry
intercepts - y - none
- x = +-2
relative extrema :
Taking f'(x) and solving for 0 to get the critical numbers
f'(x) = 3x^2 - 16/x^2 - 8
setting = to 0, gives critical numbers of +-2
Plugging in critical numbers into f(x)
f(-2) = 0
f(2) = 0
can't tell which is max or min, so make a number line, test points to left of -2, in between -2 and 2, and to the right of 2
f(-3) = -8.333
f(-2) = 0
f(-1) = -9
f(1) = 9
f(3) = 8.333
This means that a maximum is at -2,0 and minimum at 2,0
Is that correct for the relative extrema?
intervals of increasing/decreasing behavior
For the intervals, once again, i found the critical number, made the number ilne and picked test points
From (-∞ to -2) the graph was increasing, (-2 to 0) decreasing , (0 to 2) decreasing, and (2 to ∞) increasing
whats the right way to do this problem
i know we need to take the second derivative and set = to 0 but it is not working out? Are there no inflection points? and if there are no inflection points, how can there be any intervals of concavity? Can there?
f"(x) = 6x+32/x^3
cant solve for f"(x) = 0
What do i do?
intervals of concave up/down
f(0) is undefined. What does that mean about the inflection points though? I dont think there are any inflection points.. and as for concavity, i think the graph just concaves down from -∞ to 0 and concaves up from 0 to ∞..if you graph it, its just two parabolas, one opening down from -∞ to 0 and one opening up from 0 to ∞,
i sitll dont know what to do for the last two, inflection and intervals
Sorry, I don't normally use terms like "undefined",
since it doesn't really say much.
For your function, x=0 causes 16/x to shoot to infinity,
plus infinity for x>0 and minus infinity for x<0.
f(x) approaches infinity because the 16/x term is dominant for x near zero.
Your graph 'looks like' a pair of parabolas.