One of our homework problems asks us to graph a functio given several characteristics, but I don't really understand how to do it.
Suppose that a function f has the following characteristics
ii) f continuous except at x=2
iii) lim f(x) = +infinity
iv) lim f(x)=0
v) lim f(x)=3
vi) f ' exists except at x=2, -1
vii) f ' > 0 if -1 is less than or equal to x is less than or equal to 2
viii) f ' < 0 if x< -1 or x> 2
ix) f '' < 0 if x <0 and x is not equal to -1
x) f '' > 0 if x > o and x is not equal to 2
Sketch the graph of the function f.
does vi mean that there are asymptotes at those values on f? If so, how do you reconcile that with the horizontal asymptote at 0? I don't really understand how to apply any of the f ' or f '' characteristics listed to the graph of f, either.
Could someone point me in the right direction? Thanks.
vi means the derivative does not exist at x = -1 and x = 2. The x = 2 discontinuity should be obvious from iii) where there is an obvious vertical asymptote in f(x). Since f(x) is continuous at x = -1 and f'(x) is not this means the function comes to a sharp "point." An example of this is the function f(x) = |x| at x = 0.
Originally Posted by PurpleLeprechaun
An example of how to use the first and second derivatives:
Remember that f'(x) tells you what the slope does and f''(x) tells you the curvature. So start with iv) which tells us we have the horizontal asymptote y = 0 for large negative x. Then we know from viii) that the function is decreasing for x < -1. Since we also know from ix) that the function has a negative concavity for x < -1 as well. All this means that the function curves (ie. is not a line) downward from -infinity to x = -1. Since x = -1 is not a vertical asymptote, it curves down to some real value f(-1).
To give you an example of what the function might look like for (-infinity, -1] consider the function y = 1/x over the same domain.