Cracking The Ap Calculus ab Exam 2018

21. 6

Notice  that    if  we  plug    in  0   for h,  we  get  ,  which   is  indeterminate.  If  we  expand  the expression

in  the numerator,  we  get  .  This    simplifies  to   .  Next,   factor  h

out of  the top expression:  .  Now,    we  can cancel  the h   and evaluate    the limit   to  get:

    (6  +   h)  =   6   +   0   =   6.

22.

Notice  that    if  we  plug    in  0   for h,  we  get  ,  which   is  indeterminate.  If  we  combine the two

expressions on  top with    a   common  denominator,    we  get     =       =   

. We can simplify the top expression, leaving us with: . Next, simplify

the expression  into        =    .  We  can cancel  the h   to  get  .  Now,

if  we  evaluate    the limit   we  get     =   .

SOLUTIONS TO PRACTICE PROBLEM SET 2

1. Yes. It satisfies all three conditions.

In  order   for a   function    f(x)    to  be  continuous  at  a   point   x   =   c,  it  must    fulfill all three   of  the

following   conditions:

Condition   1:  f(c)    exists.

Condition   2:   f(x)   exists.

Condition   3:   f(x)   =   f(c)

Let’s   test    each    condition.

f(2)    =   9,  which   satisfies   condition   1.