165
Math 170-22: Quiz #1 (upto 2.3)
Name:
1. Suppose that P(x) is a polynomial of degree 5.
Solutions
(4 points each)
(a) How many x-intercepts can the graph have?
At most 5
(b) How many y-intercepts can the graph have?
Only 1
(c) Is it possible that P has no real roots? Justify your answer.
No be because odd degree polynomials have  for a range.
2. Find the domain of each of the following functions.
(a) f x  
(6 points each)
2x  6
x4
[-3, 4) and (4, ∞)
(b) g x   4.23  8.25 ln 2 x  7 
(-3.5, ∞)
3. True or false: The functions f x  
x 3
1
and g x  
are the same function. Explain.
2
x3
x 9
(6 points)
No because f(3) d.n.e. but g(3) = 1/6
4. For the following, sketch the graph of a function satisfying the given properties or explain why no
such function exists. (5 points each)
(a) f 2  0 but f has no local extrema at x = 2
x = 2
(b) f has a local maximum at x = 3 and f x   0 on [0, 5].
Not possible – by second derivative test, if f has a local
maximum at x = 3, then f”(3) ≤ 0
(c) f x   0 on [0, 6], f  x   0 on [0, 3], and f  x   0 on [3, 6].
x = 0
x = 3
x = 6
5. Suppose that f(0) = 2 and f x   4 for all x. What can be said about f(1) and f(-3)?
f(1) – f(0) ≤ 4(1 – 0)
f(1) – 2 ≤ 4
f(1) ≤ 6
(8 points)
f(0) – f(-3) ≤ 4(0 – (-3))
2 – f(-3) ≤ 12
-f(-3) ≤ 10
f(-3) ≥ -10
6. Use the graph of y  f  x  shown below to graphically estimate the value of f 2  .
(6 points)
using (2,2) and (1.7,0) we
get that the slope is about
6.666666…
7. Use the graph of y  g  x  shown below to answer the questions.
(5 points each)
(a) Over what intervals in [-4, 4] is g
decreasing?
[-1.3, 2]
(b) Over what intervals in [-4, 4] is
g  x   0 ?
[-4, -1.3] and [2, 4]
(c) Over what intervals in [-4, 4] is
g x   0 ?
[-4, .3]
8. Use the graph of y  f  x  shown below to answer the questions.
(5 points each)
(a) Over what intervals in [0, 8] is f concave
down?
[0, 2.8] and [4.9, 6.9]
(b) Over what intervals in [0, 8] is f
increasing?
[0, 2] and [4, 6]
(c) Where in [0, 8] does f have local
extrema? Specify the type of each.
Max at x=2 and x=6
Min at x = 4
(d) Where in [0, 8] does f have inflection
points?
X=2.8, x=4.9, x=6.9
(e) Give the equation of the tangent line to y = f(x) at the point (1.5, 8).
m = f’(1.5) which is about 0.5 from graph above
so y-8 = 0.5(x-1.5) or y = 0.5x + 7.25
9. Use the formal definition of derivative to algebraically find the derivative formula for the function
having formula f x   2 x  5 . (8 points)
f   x   lim
f  x  h  f  x
h 0
2 x  h  5  2x  5
 lim
h 0
h
h
 2  x  h   5  2 x  5  2  x  h   5  2 x  5 


 lim 
h 0 
h
 2  x  h   5  2 x  5 



2 x  2h  5   2 x  5 
 lim
h 0
h 2 x  h  5  2x  5


2
 lim
2 x  h  5  2x  5
h 0
2

2 2x  5
1

2x  5
10. Give the derivative formula for each of the following functions. Simplify as appropriate.
(7 points each)
(a) f x   3.27 x 4  7.32 x 2  11.82 x  21.25
f   x   13.08 x3  14.64 x  11.82
(b) f x   5 x 2 
7
x3
f   x   52 x  5  21x 4 
2
3
5
5 x
3

21
x4
(c) f x   5 x
f  x 
1
2
 5 x   5  
1
2
5
2 5x
11. Given f x   x 48  3x17  41x 2  5 , find a formula for f 50 x  .
(6 points)
Since the polynomial is only degree 48, the 50th derivative must be
simply f(50)(x) = 0.
12. Consider the graph of y  f  x  shown below.
(3 points each)
(a) Find lim  f x 
x 4
6
(b) Find f(2)
1
(c) Find lim f x 
x2
d.n.e.
(d) State each x value at which the function f is not continuous and give an appropriate
explanation as to why f is not continuous at each. (8 points)
x = -6 b/c limit at x = -6 does not equal f(-6)
x = -4 b/c limit does not exist at x = -4
x = -1 because f(-1) does not exist
x = 2 because limit does not exist at x = 2
13. Find each of the following limits.
(7 points each)
x 2  4 x  21
x3
x2  9
 x  3 x  7 
 x  7  10 5
 lim
 lim
 
x 3  x  3 x  3
x 3  x  3
6 3
(a) lim
2x  h
2
(b) lim
h 0
 
 5x  h   7  2 x 2  5 x  7
h

2 x 2  4 xh  2h 2  5 x  5h  7  2 x 2  5 x  7
h 0
h
 lim
 lim
h  4 x  2h  5 
h 0
-or-
d
dx
 2x
2
h

 5x  7  4x  5
 4x  5