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Math 105a Exam 1 Solutions: Derivatives and Limits, Exams of Calculus

Solutions to exam 1 of math 105a, focusing on derivatives and limits. It includes finding the greatest and least values of functions, using limit definitions to find derivatives, and analyzing the behavior of functions at discontinuities.

Typology: Exams

2012/2013

Uploaded on 03/06/2013

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Math 105a Exam 1
Solutions
1. The graph of
f
0
is provided on the left, while the graph of
f
is provided on the right.
(a)
f
is greatest at D
(b)
f
is least at A
(c)
f
0
is greatest at A
(d)
f
0
is least at C
(e)
f
00
is greatest at D
(f)
f
00
is least A
(g)
f
is increasing everywhere except C since
f
0
(
x
) = 0 at
x
=C.
(h)
f
is increasing most rapidly at A since
f
0
(
x
) is greatest there.
2. Use the limit denition of the derivative to nd
f
0
(
x
) for
f
(
x
) = 5
x
.
f
0
(
x
) = lim
h
!
0
f
(
x
+
h
)
f
(
x
)
h
= lim
h
!
0
5
x
+
h
5
x
h
= lim
h
!
0
5
x
5(
x
+
h
)
xh
(
x
+
h
)
= lim
h
!
0
5
h
xh
(
x
+
h
)= lim
h
!
0
5
x
(
x
+
h
)=
5
x
2
3. (a)
f
(
x
) = 3
x
13
x
2
1
x
+
e
2
f
0
(
x
) = 3
x
ln 3
26
x
+1
x
2
(b)
f
(
x
) = 5
x
x
3
+ 8
x
7
f
0
(
x
) = 5
x
ln 5
x
3
+ 8
x
7
+ 5
x
3
x
2
+ 56
x
6
= 5
x
x
2
8
x
5
ln 5 + 56
x
4
+
x
ln 5 + 3
(c)
f
(
x
) =
e
x
+
x
1
x
2
f
0
(
x
) = (
e
x
+
x
ln
)(1
x
2
)
(
e
x
+
x
) (
2
x
)
(1
x
2
)
2
(d)
f
(
x
) =
q
1 +
p
x
=
1 +
x
1
=
2
1
=
2
f
0
(
x
) = 1
2
1 +
x
1
=
2
1
=
2
1
2
x
1
=
2
=1
4
p
x
p
1 +
p
x
4. If
f
(
x
)=3
x
3
4
x
2
at
x
= 1, then
f
0
(
x
)=9
x
2
8
x
. At
x
= 1 the slope
m
is
f
0
(1) = 1. The line is
tangent at (1
; f
(1)) = (1
1) and
y
-intercept of the tangent line is
b
=
y
mx
=
1
(1)(1) =
2.
Therefore the equation of the tangent line is
y
=
x
2.
1
pf3

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Math 105a Exam 1 Solutions

  1. The graph of f 0 is provided on the left, while the graph of f is provided on the right.

(a) f is greatest at D (b) f is least at A (c) f 0 is greatest at A (d) f 0 is least at C (e) f 00 is greatest at D (f) f 00 is least A (g) f is increasing everywhere except C since f 0 (x) = 0 at x =C. (h) f is increasing most rapidly at A since f 0 (x) is greatest there.

  1. Use the limit de nition of the derivative to nd f 0 (x) for f (x) = x^5.

f 0 (x) = lim h! 0 f^ (x^ +^ h h)^ ^ f^ (x)= lim h! 0 x+^5 h ^ x^5 h = lim h! 0

5 x 5(x + h) xh(x + h)

= lim h! 0 xh(^ x^5 +h h) = lim h! 0 x(x^ +^5 h) = (^) x^52

  1. (a) f (x) = 3x^ 13 x^2 (^) x^1 + e^2

f 0 (x) = 3x^ ln 3 26 x + (^) x^12

(b) f (x) = 5x^

x^3 + 8x^7

f 0 (x) = 5x^ ln 5

x^3 + 8x^7

  • 5x^

3 x^2 + 56x^6

= 5xx^2

8 x^5 ln 5 + 56x^4 + x ln 5 + 3

(c) f (x) = e

x (^) + x 1 x^2 f 0 (x) = (e

x (^) + x (^) ln )(1 x (^2) ) (ex (^) + x) ( 2 x) (1 x^2 )^2 (d) f (x) =

q 1 + p x =

1 + x^1 =^2

f 0 (x) =^12

1 + x^1 =^2

 12 x^1 =^2 = 1 4

p x

p 1 + px

  1. If f (x) = 3x^3 4 x^2 at x = 1, then f 0 (x) = 9x^2 8 x. At x = 1 the slope m is f 0 (1) = 1. The line is tangent at (1; f (1)) = (1 1) and y-intercept of the tangent line is b = y mx = 1 (1)(1) = 2. Therefore the equation of the tangent line is y = x 2.
  1. Consider f (x) =

3 x^2 ; if x  0 10(x + 3) (x + 3)(x + 4) ;^ if^ x <^0

Note: (^) (x10( + 3)(x^ + 3)x + 4) = (^) x^10 + 4 provided x 6 = 3

(a) (^) xlim! 0 f (x) = (^) xlim! 0 x 10 + 4 =^104 =^52 (e) (^) x!lim 3 f (x) = (^) x!lim 3 x 10 + 4 = 10

(b) (^) xlim! 0 + f (x) = (^) xlim! 0 + 3 x^2 = 0 (f) (^) x!lim 3 + f (x) = (^) x!lim 3 + x^10 + 4 = 10

(c) (^) x!lim 4 f (x) = (^) x!lim 4 x 10 + 4 = 1 (g) f (0) = 3(0)^2 = 0

(d) (^) x!lim 4 + f (x) = (^) x!lim 4 + x^10 + 4 = 1 (h) f (3) is unde ned

Using these limits we can classify all points of discontinuity as either removable or non-removable. First notice that the function f is not continuous at x = 0, x = 3, and x = 4. In particular, (i) f has a removable discontinuity at x = 3 since (^) xlim! 3 f (x) = 10 even though f (3) is unde ned. (ii) f has a non-removable discontinuity at x = 4 since (^) x!lim 4 f (x) = 1 and (^) x!lim 4 + f (x) = 1. In other words, the line x = 4 is a vertical asymptote for the graph of f. (ii) f has a non-removable jump discontinuity at x = 0 since (^) xlim! 0 f (x) =^52 and (^) xlim! 0 f (x) = 0.

The graph is shown below.

  1. If f (x) = x^3 + 12x + 5 then f 0 (x) = 3 x^2 + 12 and f 00 (x) = 6 x.

(a) f 0 (x) = 3 x^2 + 12 = 3(x + 2)(x 2) and f 0 (x) = 0 when x = 2. Creating a sign chart shows that f is decreasing on (1; 2) and (2; 1 ).

(b) f 00 (x) = 6 x > 0 if x < 0. Therefore, f is concave up on (1; 0).