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Midterm

01:640:135 Midterm: 01:640:135 Test Fall 2016Exam


Department
Mathematics
Course Code
01:640:135
Professor
All
Study Guide
Midterm

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1A
(10) 1. Suppose f(x)=2x
2
3x. Use the denition of derivative to nd f
0
(x).
f
0
(x)=lim
h!0
f(x+h)f(x)
h
= lim
h!0
2(x+h)
2
3(x+h)(2x
2
3x)
h
= lim
h!0
2x
2
+4xh +2h
2
3x3h2x
2
+3x
h
= lim
h!0
4xh +2h
2
3h
h
= lim
h!0
4x+2h3=4x3
(9) 2. Find an equation for the line tangent to the graph of y=
p
x+2x
2
at the point where
x=1.
y=x
1=2
+2x
2
,soy
0
=
1
2
x
1=2
+4x. At x= 1, the value yis 3 and the value of y
0
is 9=2.
Thus an equation for the tangent is
y3=
9
2
(x1):
(12) 3. Assume that the functions u(x) and v(x) are dened and dierentiable for all real
numbers x. The following data is known about u,v, and their derivatives.
x u(x)v(x)u
0
(x)v
0
(x)
2 3 4 1 2
3 2 1 3 1
4 1 3 0 2
Dene f(x)=u(x)
2
+2v(x) and g(x) = v(x)=u(x). Answer the following, giving a brief
explanation of how the answers were obtained.
a) What is f
0
(2)?
Since the chain rule had not been covered when the test was given, to dierentiate u(x)
2
we have to write it as u(x)u(x) and use the product rule.
f
0
(x)=u(x)u
0
(x)+u
0
(x)u(x)+2v
0
(x)=2u(x)u
0
(x)+2v
0
(x):
Thus
f
0
(2) = 2(3)(1) + 2(2) = 2:
b) What is g
0
(3)?
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2A
g
0
(x)=
u(x)v
0
(x)v(x)u
0
(x)
u
2
(x)
Thus
g
0
(3) =
2(1) 1(3)
2
2
=
5
4
:
c) What can be said about the number and location of solutions to the equation f(x)=6:5
with xin [2;4]?
From the table, we have f(2) = 17, f(3) = 6, and f(4) = 7. By the Intermediate Value
Theorem, there is at least one solution to the equation f(x)=6:5 in the interval [2;3] and
at least one in the interval [3;4]. Thus the total number of solutions is at least 2.
(12) 4. Suppose that the function f(x)isdescribed by
f(x)=
x+Bif x<1
Ax +3 if x1
:
a) Find Aand Bso that f(x) is continuous for all numbers and f(1) = 0. Briey explain
your answer.
The only place that fmight not be continuous is at x=1,where the denition changes.
Now f(1) = A+ 3 while lim
x!1
f(x)=1+B. If fis to be continuous at x=1,wemust
have A+3=1+B.
The value of f(1) is 1+B, whichmust be 0. This gives B=1.Substituting this value
in the previous equation, we get A=1.
b) Sketch y=f(x) on the axes given for the values of Aand Bfound in a) when xis in
the interval [2;2].
–4
–3
–2
–1
1
2
3
4
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