Summary MATH 100 midterm exam review 2

MATHEMATICS 100 – Practice Problem Solutions


1. Analytically, first note that if 2 x  1  0, the inequality is automatically satisfied. So
1
part of the solution is x   . For 2 x  1  0 (so that x   12 ) square both sides and
2
gather all terms on one side of the equation:
x  1  2 x  1  x 2  2 x  1  4 x 2  4 x  1  3x 2  6 x  0  3xx  2  0. Thus,
2 2


using x  0 and x  2 to divide up the number line into intervals gives the following:




The solution satisfying x   12 is  12 ,0 .
Thus, the solution to the inequality is  ,0 .
Graphically, check out the graph below:




Again, the solution is  ,0 .

2. This is an ugly function, but you approach the question of its domain the same way
that you should approach the question of any domain: start with all values of x and
remove any values that cause any part of the function to become undefined. Examine
each part of f  x  separately.
3
1
ex is defined for all x, so no restriction to the domain from this term.

 x 
ln   is only defined provided that its argument is positive, so we need
 x  x 1 
2




1

, x
 0. First, this means that x  0, and so x  0 . Second, since the numerator
x  x2 1
is now positive, we need x  x 2  1  0 as well. For this quantity to even be defined, we
must have x 2  1  0, which means x must satisfy either x  1 or x  1 . Finally, we
also need x  x 2  1  0. Note that for this to be true, x must be positive , since

x 2  1  0. So, for x  0, we have:
x  x 2  1  0  x  x 2  1  x 2  x 2  1,
which is always true.
Thus, for this part of the function f  x  to be defined, we must have x  1 .

arcsin  e x  is defined only when its argument is between -1 and +1. Since e x  0, this
means we must have e x  1  x  ln1  x  0 .
Sooooo, for all parts of f  x  to be defined we need both x  1 and x  0. This is clearly
impossible, so the domain of f is  (the empty set). (Interesting, eh? This function is
undefined for all x.)

x
3. a) We must have  0, since you can only take the ln of positive numbers. To
x 1
solve this inequality: set the top to 0 (giving x  0 ), set the bottom to 0 (giving x  1 ).
x
These numbers then divide the number line into intervals upon which is always of
x 1
one sign. Pick a point in each interval and sub it into the function:
x
 , 1 : sub x  2 to find that 0
x 1
x
 1,0  : sub x   12 to find that 0
x 1
x
 0,   : sub x  1 to find that 0
x 1
The domain of f is thus  , 1   0,   .
 x 
b) Interchange x and y in the formula y  ln   and solve for y:
 x 1 
 y  ex
x  ln    e x

y
  
y  1 e x
 y  e x
 y  ye x
 y 1 e x
   y.
 y 1  y 1 1 ex
ex
Thus, f 1  x   . Since e x is defined for all x, the only restriction to the domain is
1 e x


that 1  e  0  x  0. The domain of f 1  x  is thus  ,0    0,   .
x


c) Since f and f 1 are inverse functions, the domain of one is the range of the other.
Thus, the range of f is  ,0    0,   , and the range of f 1 is  , 1   0,   .

2

, 4. y  e x , x  1, 0  y  e. Interchange x and y to obtain:
x  e y , y  1, 0  x  e  y  ln x, y  1, 0  x  e.
y   x, x  1, y  1. Again interchange x and y:
x   y, y  1, x  1  y   x, y  1, x  1.
The inverse function is therefore:
ln x, 0  x  e
f 1  x    .
 x, x  1
Its domain is thus  , 1   0, e.

5. i) First, the  1n causes the a n to flip signs; thus, the limit can only exist if the
remaining part of the term goes to zero. Since the higher power of n is in the
denominator, you should expect this to be the case. To prove that it is:
n 1

  0  lim  1 4
n 0 n
 lim n 4 1  lim
n4 n3
 0.
n
lim 4
n  n  1 n  1  
4 
n  1
1 n  n 1
4 4
n n n

ii) This has the form   , which is indeterminate. This means you have to simplify.
Because of the cube roots, recall the following: a 3  b 3  a  ba 2  ab  b 2 . Setting
a  3 n 3  n 2  1 and b  3 n 3  1 gives the following:
n 3
 
 n 2  1  n3  1    3
n3  n2  1  3 n3  1 n3  n2  1  2/3

 n3  n2  1  n
1/ 3 3
1 
1/ 3
 
 n3  1
2/


which simplifies to:
n2  n
3 3
 n 2  1  3 n3  1 n3  n 2  1  
2/3

 n3  n 2  1  n
1/ 3 3
1 1/ 3

 n3  1 
2/3

or
n2
n  n 1
3 2

2/3

 n  n 1
3 2
 n
1/ 3 3
1 
1/ 3

 n 1 3
 2/3
  3 n 3  n 2  1  3 n 3  1.

Thus,
n2
lim 3 n 3  n 2  1  3 n 3  1  lim
n  n 
n 3
 n2 1 
2/3

 n3  n 2  1  n
1/ 3 3
1 
1/ 3

 n3  1  2/3

n2

lim
n 
n 1 
2 1
n  n13 2/3

 n 2 1  1n  n13  1  
1/ 3 1 1/ 3
n3

 n 2 1  n13  
2/3


1 1
 .
lim
n 
1  1
n  n3

1 2/3

 1  1n   1  
1 1/ 3
n3
1 1/ 3
n3
 1  
1 2/3
n3
3

6. a) This means to check whether or not the limit as n   exists. For functions
involving powers of n, factor out the highest power of n from the numerator and
denominator:
3
6n 8  3n 2  cos n 3 n 8 6  3  cos n
n6 n8

n  3 6  n36  cos
n8
n

 lim  lim .
 
lim
n 4
n 9  5n 3  sin n n 4
n 9 1  56  sin9n n
n  4 1  n56  sinn9n
n n

8 9 5
Note:   ; thus we have
3 4 12
3