false

Study Guides
(248,228)

Canada
(121,418)

Western University
(12,906)

Calculus
(44)

Calculus 1000A/B
(34)

Chris Brandl
(3)

Quiz

Unlock Document

Calculus

Calculus 1000A/B

Chris Brandl

Fall

Description

MATH 127 Fall 2012
Assignment 3 Solutions
1. Solve the following equations for x.
q
▯x 2x 5 x
(a) e 3e ▯ 4 = e
r
3e2x▯ 5 = e2x
4
5
3e2x▯ = e4x
4
5
e4x▯ 3e2x+ = 0
4
2x 2 2x 5
(e ) ▯ 3e + = 0
4
Notice that this is like a quadratic with respect tope . Using the quadratic
3 ▯ 9 ▯ 5 3 ▯ 2 5 1
formula we get that the roots are given by ex = = = ; .
2 2 2 2
2x 5
First consider e = 2.
5
e2x=
2 ▯ ▯
5
2x = ln
2
▯5▯
x = ln 2
2
▯1▯
Now consider e2x = 1 . Solving for x we get x =n 2 .
2 2
▯5▯ ▯ 1
Thus, the two solutions to the equation are x =n 2 and x = ln 2 .
2 2
(b) ln(2ln(x)) = 3
ln(2ln(x)) = 3
3
2ln(x) = e
3
ln(x) = e
2
e
x = e 2
1 (c) ln(x ▯ 2e) + ln(x ▯ 3e) = 2
We ▯rst simplify the left hand side using properties of logarithms.
ln(x ▯ 2e) + ln(x ▯ 3e) = ln((x ▯ 2e)(x ▯ 3e))
2 2
= ln(x ▯ 5ex + 6e )
Thus, we have that ln(x ▯ 5ex + 6e ) = 2. From this we get x ▯5ex+6e = e 2 2 2
2 2
which in turp gives x ▯ 5ex + 5e = 0. Tpe quadratic formula gives that
2 2
5e ▯ 25e ▯ 20e (5 ▯ 5)e
x = = . Now since we need x > 3e for
2 2 p
(5 + 5)e
ln(x ▯ 2e) + ln(x ▯ 3e) to be de▯ned we have the single solution x = .
2
2. Let f(x) = ▯e ▯x + 2.
(a) Sketch the graph of f, labelling any asymptotes and x and y-intercepts.
3
2.5
2
y=2
1.5
1 (0,1)
0.5
(-ln(2),0)
-4 -3.5-3-2.5-2-1.5-1-0.50 0.5 1 1.5 2 2.5 3 3.54
-0.5
-1
-1.5
-2
-2.5
-3
▯1
(b) Explain why the inverse exists based on the graph. Find f .
We can see that the graph is strictly increasing and so, passes the horizontal line
test. That is, we see that the function is one-to-one and from this we get that the
2 inverse exists. Solving for the inverse we get
y = ▯e ▯x + 2
▯x
y ▯ 2 = ▯e
▯x
2 ▯ y = e
ln(2 ▯ y) = ▯x
x = ▯ln(2 ▯ y)
▯1
f (x) = ▯ln(2 ▯ x)
(c) Sketch f ▯1 and label the x and y-intercepts on the same axes used in part a.
3
2.5
y=2 2
1.5
1 (0,1)
0.5
(-ln(2),0)
-4 -3.5-3 -2.5-2-1.5-1 -0.50 0.5 1 1.5 2 2.5 3 3.5 4
-0.5 (1,0)
x=2
(0,-ln(2))
-1
-1.5
-2
-2.5
-3
3. For each of the following functions, state the domain of the function, ▯nd its inverse,
and state the domain of the inverse.
p
x
(a) f(x) = 2 ▯ e

More
Less
Related notes for Calculus 1000A/B

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.