Math 105a Solutions to Review Problems for Exam 1
x2 + 5x + 6
(x + 2)(x + 3)
= lim
= lim x + 3 = 1
x! 2
x! 2
x! 2
x+2
x+2
1. (a) lim
2x2 6x
2x2 6x
2x(x 3)
= lim
= lim
= lim 2x = 6
(x 3) x!3
(x 3) x!3
x! 3 j x 3j
x !3
Note: if x ! 3 then x 3 < 0, therefore, jx 3j = (x 3).
(b) lim
x4
= lim
(c) lim 3
x!1
x!1 x + 1
x+4
=1
(d) lim+
x! 2 x 2
x3
x3
x4
x3
+ x3
1
= lim
x!1
x
=1
1 + x13
2. Suppose lim f (x) = 2 and lim g (x) = 3. Then
x! 4
x! 4
(a) lim (g (x) + 3) = lim g (x) + lim 3 = 3 + 3 = 0
x! 4
x!4
(b) lim xf (x) = lim x
x! 4
x !4
x !4
lim f (x) = 4 2 = 8
x !4
f (x)
limx!4 f (x)
2
1
(c) lim
=
=
=
x!4 g (x) 1
limx!4 g (x) limx!4 1
3 1
2
3. Recall that the average velocity of the particle is the slope of the line passing through the points
on the curve of f when t = 1 and t = 5. Similarly, the instantaneous velocity is the slope of the
tangent line to the curve at t = 3. A graph is shown below.
4. Consider the graph of f given below. Sketch a graph of f 0 .
5. Below is a graph of a function where lim f (x) = 2 and lim f (x) = 4.
x!1
x!
1
1
6. Below is a graph of a function where lim f (x) = 5 and f (3) = 1.
x!3
8
2
>
<x + 1;
if x < 0
7. Let f (x) = 1;
if x = 0 :
>
:
cos x; if x > 0
The graphs of f and f 0 are shown below.
8. (a) A function f is continuous at a point x = a if lim f (x) = f (a). In other words, as x ! a
x !a
the left and right hand limits are equal to each other and equal the value of the function at
x = a.
(b) Graphical interpretation of a removable discontinuity: there is a hole in the graph of f when
x = a. This means that lim f (x) exists but doesn't equal f (a). The discontinuity is removx!a
able because we could rene the function at x = a to ll in the hole in the graph in order to
make it continuous.
Graphical interpretation of a non-removable discontinuity: there is a break or jump in the
graph of f when x = a. This happens when lim f (x) does not exist because the left and right
x!a
hand limits dier. In this case, there is no way to redene function at x = a to repair the
break or jump.
(c) Below is a graph of a function with a removable discontinuity.
(d) Below are the graphs of two functions, each with a non-removable discontinuity.
2
x2 + 5x + 6 (x + 2)(x + 3) x + 3
=
=
provided x 6= 2.
x2 x 6
(x + 2)(x 3) x 3
The function f is not continuous at x = 2 and x = 3. In particular, f has a removable discon1
tinuity at x = 2 since lim f (x) =
even though f ( 2) is undened. The function has a
x! 2
5
non-removable discontinuity at x = 3 since lim f (x) = 1 and lim+ f (x) = 1. In other words,
x!3
x !3
the line x = 3 a vertical asymptote for the graph of f . The graph is shown below.
9. f (x) =
10. Using the limit denition of derivative to nd f 0 (x):
p
(a) f (x) = 2 x
p
p
p
px)
2 x+h 2 x
2( x + h
f 0 (x) = lim
= lim
h!0
h!0
h
h
p
px) px + h + px
p
p
2( x + h
= lim
h!0
h
x+h+ x
2(x + h x)
2
2h
2
p
= lim p
= lim p
= p
p
p
p
h!0 h( x + h + x)
h!0 h( x + h + x)
h!0 x + h + x
2 x
= lim
f 0 (x) =
(b) f (x) =
p1x
1
x3
1
(x+h)
3
f 0 (x) = lim
h !0
h
h !0
1
12. (a) f (x) = 5x2 +
f 0 (x) = 10x
p
(b) f (x) = 3 4x7
f 0 (x) =
sin( 2 + h)
h!0
h
= lim
p3
1
x2
h!0
h
= lim
x3
h!0
3
2
4 1=3
x = 10x
3
3
3x2 = 4x7
1
4x7
3
(c) f (x) = 3 2x + x3
3x2
sin( 2 )
2 = 5x2 + x
x4
2x
2
3
(x +3x h+3xh +h )
x (x+h)
3
3
h
3
x4
f 0 (x) =
sin( 2 + h)
h!0
h
= lim
x3 (x+h)3
x3 (x+h)3
h(3x2 + 3xh + h2 )
(3x2 + 3xh + h2 )
3x2
= lim
= 6 =
3
3
3
3
h!0
hx (x + h)
x ( x + h)
x
= lim
11. lim
1
x3
2=3
3x2
1=3
(28x6
= f0
2
2
x3
6x) =
2
f 0 (x) = 3(ln 2)2x + 3x2
3
2
x4=3
p
where f (x) = sin x.
2
43x
3
28x6 6x
3 (4x7 3x2 )2=3
3
x4
2x + ex
2x
((ln 2)2x + ex ) 2x (2x + ex ) (ln 2)2x ex (1 ln 2)
=
f 0 (x) =
(2x )2
2x
(d) f (x) =
(e) f (x) = (x
5)ex
f 0 (x) = 1 ex
(f) f (x) =
2 +1
+ (x
5)ex
2 +1
2 +1
(2x) = ex
(2x2
10x + 1)
35x 1
(3x + 1)2
f 0 (x) =
=
13. f (x) =
2 +1
(5 ln 3)35x 1 (3x + 1)2 35x 1 (6)(3x + 1) 35x 1 (3x + 1) [(5 ln 3)(3x + 1)
=
(3x + 1)4
(3x + 1)4
35x
1
(15x ln 3 + 5 ln 3
(3x + 1)3
6)
p
1
x2 + 9 = (x2 + 9)1=2 , therefore f 0 (x) = (x2 + 9) 1=2 2x =
2
f ( 4) = 5 and f 0 ( 4) = 4=5.
p
x
x +9
2
Using slope m = 4=5 and the point ( 4; 5) we nd the equation of the tangent line:
y = 4=5x + 9=5.
14. Given: y = f (x) with f (1) = 8 and f 0 (1) = 3.
p
(a) g (x) = 3 f (x) = [f (x)]1=3
1
therefore g 0 (x) = [f (x)] 2=3 f 0 (x) and
3
1
2=3
0
g (1) = [f (1)]
f 0 (1) = 13 (8) 2=3 ( 3) =
3
p
(b) g (x) = f ( x) = f (x1=2 )
p 1
therefore g 0 (x) = f 0 ( x) x 1=2 and
2
p 1 1=2 1 0
8
0
0
g (1) = f ( 1) (1)
= f (1) = = 4.
2
2
2
4
1
4
6]
2
15. Consider f (x) = x3 e x .
2
2
(a) f 0 (x) = 3x2 e x + x3 e x ( 2x) = (3x2
f 00 (x) = (6x
2
8x3 )e x + (3x2
2
2x4 )e x and
2
2x4 )e x ( 2x) = 2x(2x4
2
7x2 + 3)e x .
r
p
2
3
6
(b) First notice that f 0 (x) = x2 e x (3 2x2 ) = 0 when x = =
and x = 0. Use a sign
2
2
chart to determine where the rst derivative is positive and negative.
f increasing on (
f decreasing on (
p
p
6=2;p0) and (0; p6=2).
1; 6=2) and ( 6=2; 1).
2
2
(c) First notice
f 00 (x) = 2x(2x4 7x2 + 3)e x = 2x(2x2 1)(x2 3)e x = 0 when
p that
p
2
x = ; 3; and 0. Use a sign chart to determine where the second derivative is positive
2
and negative.
p
p
p
p
f is concave down on p
( 1; p 3), ( p2=2; 0), and p( 2=2; 3).
f is concave up on ( 3;
2=2), (0; 2=2), and ( 3; 1).
(d) lim f (x) = 0 and lim f (x) = 0. (You may need to construct a table to see this.)
x!1
x!
1
(e) Below is a sketch of the graph of f .
5