Math 131 Exam 1 Solution Fall 1999

1. (25 points) a)

b) Support your result in part (a) both by a table and by a graph of the function in a window containing the point . Notice that .

2. a) The function is continuous at the point x=2 if and only if the two one sided limits and are equal. We determine the value of k for which the function is continuous by

i) calculating each of the one sided limits in terms of k.

ii) set them equal to each other

and solve for k to get k=2.

b) The function is not differentiable even when k=2 because we have a corner at the point (2,6) on the graph. This can be seen by comparing the derivatives at the point x=2 of the two functions 2x+2 and (which are not equal). More precisely, we calculate the left hand and right hand derivative limits:

Since they are not equal, the derivative limit does not exist.

3. a) A line x=a is a vertical asymptotes of the function

if and only if a one sided limit is infinite

The candidates for vertical asymptotes are the roots x=1 and x=-4 of the denominator. We need, however, to check that a one sided limit is indeed infinite (In question 1, for example, 2 is a root of the denominator, but x=2 is not a vertical asymptote).

In our case, x=1 is a vertical asymptote because the numerator does not vanish at x=1. (If you stopped here you got the full credit). Similarly, x=-4 is a vertical asymptote because the numerator does not vanish at x=-4. We can show the vertical asymptotes by a graph: A precise answer would include a calculation of the one sided limits algebraically:

Similarly,

b) The line y=k is a horizontal asymptote of the function because

Thus, for the line y=3 to be a horizontal asymptote, we choose k=3.

4. We calculate the derivative of using the definition of the derivative as follows:

5. Let Using the derivative formula for a quotient,

we compute:

6.
a) The derivative of is

b) The equation of the tangent line through the point (1,1+e) is

Using part a) we get that . So, the tangent line is

7.
a) Using the product rule we see that, if , then

b) If g'(0)=5, then

so f(0)=5.