Key to the Practice Midterm

1.  A.  False , the limit has nothing to do with the value at the point.

B.  False.  Since lim from the left is 2 and the lim from the right is 1.  no limit implies not continuous.

C.  False.  The function must also be differentiable on (a,b) and continuous on [a,b].

2.  A.  -1.5

B.  0

C.  .25

D.  DNE

3.  The function is continuous since it is a polynomial.  f(-2) = -28 < 0 and f(1) = 5 > 0 by the intermediate value theorem, there is a c between -2 and 1 with f(c) = 0.

4.  

A)  i.  4, ii.  DNE, iii.  1, iv.  3 v. 2

B)  x = -3, -1, and 1

C)  x = -3, -1, 1, and 3

5.  Not done on the web

6.  We write a table of different  small h's for  the difference quotient [(-2 + h - 1)/(-2 + h + 1) - 3]/h

h	Difference Quotient
-.01	1.98
-.001	1.998
.01	2.02
.001	2.002
.0001	2.0002

We see that the slope of the tangent line is approximately 2.  The point is (-2,y(-2)) = (-2,3)

so the equation is  y - 3 = 2(x - (-2))

y - 3 = 2x + 4

y = 2x + 7

7. A.   2xcosx - x²sinx 

B.  1/2 x^-1/2 +3/x² + 7/6 x^1/6

C.  [(1 - x²)cosx - 2xsinx]/(1 - x²)²   

8.

A.   lim h-> 0 [(x + h)² - 2(x + h) + 4 - (x² - 2x + 4)]/h

= lim h -> 0 [x² + 2xh + h²  - 2x - 2h + 4 - x² + 2x - 4]/h

= lim h -> 0 [2xh + h² - 2h]/h

= lim h -> 0 (2x + h - 2) = 2x - 2

C.  Let e = 1 and suppose L and very small delta are given.  

If L < 0 then if x = 1 + delta/2, |1 +delta/2 - 1|  < delta while

|f(1 + delta/2) - L| = |3(1 + delta/2) + 1 - L| = |4 + 3delta/2 - L|  > 4 > e

If L > 0 then if x = 1 - delta/2, |1 - delta/2 - 1|  < delta while

|f(1 - delta/2) - L| = |2(1 - delta/2)  - 4  - L| = |-2 - delta - L|  > 2 > e