MATH 1A/1B REVIEW PROBLEMS Solution.PDF

MATH 1A/1B REVIEW PROBLEMS Solution

DIRECTIONS: In each problem below answer TRUE if the statement is always
true and FALSE otherwise. Enter your answers on a SCANTRON form 882E or
any SCANTRON form with 30 or more questions on it. Our bookstore has plenty
of these. Submit your SCANTRON form on the day listed on the course calendar.
1. d
dx
cos3
x = 3cos2
x
2. If
lim
x!3 f (x) = 7
and
g(3) = 4 , then
lim
x!3
( f (x)g(x)) = 28 .
3. If
f (x)
is differentiable and concave up for all x, then
f !(a) < f (b) " f (a)
b " a
; a < b
4. If
f (x)
is continuous at
x = a
then
f (x)
is differentiable at
x = a .
5. If
f !(x)
is increasing, then
f (x)
is also increasing.
6. If
f !(x)
is defined for all x, then
f (x)
is defined for all x.
7. If
f (x)
has an inverse function
g(x)
such that
f (x)
and
g(x)
are both
differentiable, then
g
!(x) = 1 / f !(x).
8.
d
dx
x
3
! cos x
tan x
"
#
$
%
&
' =
3x
2 + sin x
sec2
x
9.
lim
x!0
sin(2x)
x
= 2
10. If y is a differentiable function of x and sin(xy) + ln x = arctan y , then
dy
dx
=
y cos(xy) + 1 / x
1
y
2 + 1
! x cos(xy)
11. Let f and g be two functions whose second derivatives are defined. Then
( fg)!!
= f !!
" g + f " g
!!
12. If f !(p) = 0 then f (x) has either a global min or a global max at x = p .
13. If the radius of a circle is increasing at a constant rate, then so is the area.
2
14. Suppose
P(x)
is a polynomial of degree n. Then
P(n)
(x) = 0 . (Note that
P(n)
(x)
is the nth derivative of
P(x)
)
15. If
f !
is undefined at x = p, then f cannot have a local maximum or
minimum at x = p.
16. If a car decelerates at a constant rate from 10 m/s to 6 m/s in 8 seconds,
then the distance traveled during this time is 64 m.
17. If
F!(x) = x
and
F(4) = –5 , then
F(–3) = –17.5 .
18. Both the Trapezoidal and Left-Hand Sum approximations to I = –2! x dx
0
4
"
will overestimate
I .
19. Suppose that a 0. Then
1
x p
dx
0
a
! converges for p < 1 and
1
x p
dx
a
!
"
converges for p 1.
20. Every continuous function is the derivative of some function.
21. f (x)! g(x) dx = " f (x) dx (" ) g(x) dx (" )
22.
te
t dt
4
x
! = xe
x
– e
x
" 3e
4
23. If f is a continuous function and
f (x) dx
c
t
! = G(t), then G!(t) = f (t).
24.
4m – 3
(2m
2
– 3m + 2) ! 3 dm =
1
"2(2m
2
– 3m + 2)
2 + C
25. The initial value problem
dy
dx
=
x + 1
x
, y(1) = a
has solution
y = x + ln x ! 1+ a
26. If, f is continuous and positive for x a and
lim
x!"
f (x) = 0 , then
f (x) dx
a
!
"
converges.
27. There exists a function F such that
F'(x) = e
x
2
.
MATH 1A/1B REVIEW PROBLEMS
Over, please 3
28. Suppose that
f (x)
and
g(x)
are differentiable functions. Then,
f !(x)g(x) + f (x)g
!(x)
f (x)" g(x)
#
$
%
&
' ) ( dx = ln f (x)" g(x) + C
29.
1
g(x) ! dx = ln g(x) + C
30. Assume that f is a twice-differentiable function. Then integration by parts
will show that, x ! f ""(x) dx
0
b
# = b ! f "(b) – f (b) + f (0)
Use the table below to record your answers as you work through this form. Be
sure to ENTER your final answers on a SCANTRON form and submit the
SCANTRON on the day listed on the course calendar.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
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