Quiz 5
Name:
; 39. Graph the function f !x" ! x $ s$ x $ . Zoom in repeatedly,
y
; 40. Zoom in toward the points (1, 0), (0, 1), and (!1, 0) on
0
first toward the point (!1, 0) and then toward the origin.
What isMath
different
the behavior
of fI in the vicinity of
1310about
- Engineering
Calculus
September 19, 2014
these two points? What do you conclude about the differentiability of f ?
Quiz Score:
/10
Answer each questionthe
completely
area below.
all 1" . What
and explain
graph ofinthethefunction
do youyour reasoning. If
t!x" ! !x !
the work is at all ambiguous,
it
is
considered
incorrect.
No
phones,
calculators,
or notes are allowed.
notice? Account for what you see in terms of the differenAnyone found violatingtiability
these rules
of t.will be asked to leave immediately. Point values are in the square
to the left of the question. If there are any other issues, please ask the instructor.
Show2
4
2#3
work
; 45– 46 Use the defin
41. The figure shows the graphs of f , f #, and f ". Identify each
0
00
Then graph f , f #, a
curve,
yourbelow.
choices.
1. The graphs of f (x), f (x)
and fand
(x)explain
are shown
Identify each curve and explain your choices.
your answers are re
y
a
45. f !x" ! 3x 2 $ 2
b
2
; 47. If f !x" ! 2x !
x
c
Graph f , f #, f "
graphs consisten
derivatives?
48. (a) The graph o
s is measure
Solution: Consider labeling the following points:
x2
y
a
b
x1
c
x3
x
We can base our answer by referencing these points. Let us first examine x2 . We see that b is roughly
flat around x2 , meaning its derivative must be 0. Immediately, this identifies c as a candidate. We
can verify this by examining whether b is increasing or decreasing and how that relates to its derivative.
Particularly, to the left of x2 , b is increasing, meaning its derivative must be positive. To the right of x2 ,
b is decreasing, meaning its derivative must be negative. The graph c does indeed satisfy this, meaning
b0 = c.
Next, let us examine c. We may think that around x1 , c is flat so its derivative must be 0, suggesting
perhaps a as a candidate. Notice, though, to the left of x1 , c is increasing, but a is in fact, negative,
eliminating a as a possible candidate for c 0 , establishing f 00 = c, by a process of elmination.
We can also see this by directly analyzing a. Notice around x3 , a is flat meaning its derivative must be
0, suggesting b as a candidate. To the left of x3 , a is always increasing, meaning its derivative must
always be positive, which b is. To the right of x3 , a is decreasing slightly, which matches b going slightly
negative. Thus, a0 = b, soldifying our belief that f = a, f 0 = b, f 00 = c.
1/2
Quiz 5
2
Math 1310 - Engineering Calculus I
September 19, 2014
2. True or false? If true, explain why. If it is false, provide a counterexample (a drawing suffices).
(a) If f (x) is discontinuous at x = a, then f (x) is not differentiable at x = a.
Solution: We have a theorem in class that says if a function is differentiable, it is continuous. In
logical symbols, this means that D (differentiable) =⇒ (implies) C (continuous). That is, if D is
true, then C is true. The contrapositive says that if D =⇒ C, then {not C} =⇒ {not D}. That
is, if a function is not continuous, it is not differentiable, which is exactly this statement, meaning
it must be true.
Geometrically, if a function is discontinuous, what would it even mean to have a tangent line there?
What point would the tangent line go through? It doesn’t make sense to consider the tangent line
to a discontinuous point, so this further solidfies our answer.
(b) If a function f (x) is continuous at x = a, then f (x) is always differentiable at x = a.
Solution: We have seen in class if we have a corner or a vertical slope, f 0 (x) does not exist, but
neither of these conditions violates continuity, providing counterexamples to the statement. Thus,
the statement is, in general, false.
4
3. Compute the derivative of the following function using the limit definition of the derivative:
f (x) = x 2 + x.
Solution: We know the definition of f 0 (x) = limh→0
f (x+h)−f (x)
.
h
Applying this:
f (x + h) − f (x)
h
h→0
(x + h)2 + (x + h) − x 2 − x
= lim
h
h→0
2
2
x + 2xh + h + x + h − x 2 − x
= lim
h
h→0
2
2
x + 2xh + h + x + h − x 2 −
x
= lim
h
h→0
h(2x + h + 1)
= lim
h
h→0
h (2x + h + 1)
= lim
h→0
h
= lim 2x + h + 1
f 0 (x) = lim
h→0
= 2x + 1.
Note, we can easily verify this with the power rule that we recently learned, which states
which immediately provides us the same answer.
d
n
dx (x )
= nx n−1 ,
2/2