Problem Points Score
1
16
2
16
3ab
18
3cd
18
4
18
5
14
∑
100
M161, Test 1, Fall 2007
Name:
Section:
Instructor:
You may not use calculators on this exam
cos2 θ =
1+cos 2θ
2
sin2 θ =
1−cos 2θ
2
Theorem (The Derivative Rule for Inverses) If f has an interval I as domain and f 0 (x) exists and
is never zero on I, then f −1 is differentiable at every point in its domain. The value of ( f −1 )0
at a point b in the domain of f −1 is the reciprocal of the value of f 0 at the point a = f −1 (b):
1
.
( f −1 )0 (b) = 0 −1
f ( f (b))
1) Let f (x) = xex for any nonnegative real number x ≥ 0.
a) What is the range of f ?
b) Explain why f has an inverse. (If you refer to a criterion or test, indicate why the test applies!)
c) Find a value a such that f (a) = e.
d) Let g(x) = f −1 (x) the inverse of f . Determine g0 (e) = ( f −1 )0 (e).
2)
0
a) Place ln(10) approximately on this line: (remember that e ≈ 2.71828)
1
2
3
4
b) Calculate the derivative
5
6
d
ln(x4 · ex ).
dx
c) Evaluate sec(asin(5/7)).
7
8
3) Z Evaluate the following integrals. Show your work.
ex
a)
dx
1 + e2x
Z
b)
2x dx
Z
sin(x)
dx
cos(x)
Z
6
p
dx
4 − (x + 1)2
c)
d)
4) The table below gives three differential equations. For each equation, indicate which of the
four possible general solutions I-IV (i.e. one is not used) is a general solution for this equation.
(You do not need to solve the equations, just to verify which of the functions is a solution!) Furthermore, we list for each equation one initial value setting. Find the correct value for C in the
respective general solution that will solve this given initial value problem.
dy
+ 3y = ex
dx
dy
+ ex = x2
dx
dy
+ 3y = e−x
dx
y(0) = 0
y(0) = 2
y(0) = 3/2
General solution
Initial Value Problem
Value of C to solve the IVP
Possible General Solutions:
I
II
III
IV
y(x) = 1/2 e2 x +C e−3 x
y (x) = 1/4 ex +Ce−3 x
y (x) = −1/2 x2 + ex +C
y (x) = −ex + 1/3 x3 +C
Hint: The correct values for C are to be chosen from the set {±1/4, 0, ±1, ±2, ±3}, no two of
the initial value problems have the same C-value.
5) A radioactive substance has a half-life of 100 years (that is, after 100 years of decay, only half
the amount of the substance remains). How long will it take until only 10% of the initial amount
remains? (As you don’t have a calculator the result can be an expression and does not need to be
an explicit number.)