Question

Let q(x) = e^-x / X-C
be a one-parameter family of functions where c > 0.
(a) Explain why q has a vertical asymptote at x = c.
b) Determine as x goes to infinity limit q(x) and as x goes to negative infinity limit q(x).
(c) Compute q'(x) and find all critical values of q.
(d) Construct a first derivative sign chart for q and determine whether each critical value
results in a local minimum, local maximum, or neither.
(e) Sketch a typical member of this family of functions with important behaviors clearly
labeled.

Answer 1

b) I think for part b) as x-> \[x->\infty \it is positive \]

Answer 2

as x goes to negative infinity it is zero and there is no y-intercept

Answer 3

For part a, there is a vertical asymptote at x=c because that makes division by 0.

Answer 4

did I answer to part b correctly?

Answer 5

for part c0 q'(x)= - e^-x . (x-c+1) / (x-c)^2 is that right? but what are the critical values?

Answer 6

can you help me with the graph? I don't know how to do it

Answer 7

As x goes to infinty, it seems that e^-x approaches zero while x-c gets large. So, the limit seems to be 0.

Answer 8

But, I must sign off now. Sorry. I hope the others will help!

Answer 9

here is graph to give you an idea of what it looks like

http://www.wolframalpha.com/input/?i=plot+e%5E%28-x%29%2F%28x+-1%29

Answer 10

Thank you

Answer 11

as far as critical values
set derivative equal to zero
multiply out the denominator --> it goes away

\[-e^{-x}(x-C+1) = 0\]


x = C -1

Answer 12

so for part d) there is a local max right?

Answer 13

yes at C-1