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**preview**shows pages 1-2. to view the full**7 pages of the document.**Solutions to Review Sheet for Exam 1, Math 151, Fall 2017

These problems are presented in order to help you understand the material that is listed

prior to the rst exam in the syllabus. DO NOT assume that your rst midterm exam will

resemble this set of problems. The following 20 problems are not meant to be a sample

exam. These problems are just a study aid. Since we have not covered L'H^opital's Rule

yet, this rule should not be used to answer any of these questions.

(1) Simplify sin1(sin(7=4)) and sin1(sin(5=6)).

Recall that the formula sin1(sin(x)) = xis valid only when =2x=2.

sin(7=4) = sin(7=42) = sin(=4) and =2<=4< =2. Therefore,

sin1(sin(7=4)) = sin1(sin(=4)) = =4:

sin(5=6) = sin(5=6) = sin(=6) = sin(=6) and =2< =6< =2. Therefore,

sin1(sin(5=6)) = sin1(sin(=6)) = =6:

(2) Assume x0has the properties sin(x0) = 0:7 and cos(x0)<0. Evaluate sin(4x0) and

cos(4x0).

We know cos(x0) = q1sin2(x0) = p1(sin(x0))2=p1(0:7)2=p0:51 and

cos(x0) is negative. Therefore, the correct choice of is and cos(x0) = p0:51. Now

sin(2x0) = 2 sin(x0) cos(x0) = 2(0:7)(p0:51) = (1:4)p0:51

and

cos(2x0) = cos2(x0)sin2(x0)=0:51 (0:7)2= 0:02:

Then

sin(4x0) = 2 sin(2x0) cos(2x0) = 2((1:4)p0:51)(0:02) = 0:056p0:51

and

cos(4x0) = cos2(2x0)sin2(2x0) = (0:02)2((1:4)p0:51)2

= 0:0004 (1:96)(0:51) = 0:9992:

(3) Simplify cot(sin1x) and cos(tan1x).

We know (cos(sin1x))2= cos2(sin1x)=1sin2(sin1x)=1(sin(sin1x))2= 1 x2,

hence cos(sin1x) = p1x2. Since sin1xis in the interval [=2; =2], we must have

cos(sin1x)0, and this implies that the must be +. This says cos(sin1x) = p1x2.

We also know sin(sin1x) = x. This implies

cot(sin1x) = cos(sin1x)

sin(sin1x)=

p1x2

x:

1

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

We know (sec(tan1x))2= sec2(tan1x) = 1 + tan2(tan1x) = 1 + (tan(tan1x))2= 1 +

x2, hence sec(tan1x) = p1 + x2. Since tan1xis in the interval (=2; =2), we must

have sec(tan1x)>0, and this implies that the must be +. This says sec(tan1x) =

p1 + x2. This implies

cos(tan1x) = 1

sec(tan1x)=1

p1 + x2:

(4) Find all solutions of the equation

e6+x2=22

e7x= 1.

Since (et)2=e2t, this can rewritten as e2(6+x2=2)

e7x= 1. This equation is equivalent to

e2(6+x2=2) =e7x. Since eu=evimplies u=v, we now know 2(6 + x2=2) = 7x, which is

x27x+ 12 = 0. The quadratic equation gives us the solutions x= 3 and x= 4.

(5) Solve for xin the equation log27 x= 1=3.

From log27 x= 1=3 we get 27log27 x= 271=3= 3. Since 27log27 x=x, we get x= 3.

(6) The position of a particle at time tseconds is s=t2+ 1=t meters. Find the average

velocity over the time interval [1;5].

The average velocity over the time interval [1;5] is

(52+ 1=5) (12+ 1=1)

51= 5:8 meters=second:

(7) Find lim

x!5+

x5

jx5jand lim

x!5

x5

jx5j.

If xapproaches 5 from the right then x > 5 and x5>0, which implies jx5j=x5.

This gives

lim

x!5+

x5

jx5j= lim

x!5+

x5

x5= lim

x!5+1 = 1:

If xapproaches 5 from the left then x < 5 and x5<0, which implies jx5j=(x5).

This gives

lim

x!5

x5

jx5j= lim

x!5

x5

(x5) = lim

x!51 = 1:

(8) Find lim

x!4+

10 x2

(x4)5and lim

x!4

10 x2

(x4)5.

As xapproaches 4 from the right, x4 approaches 0 and remains positive. This implies

that (x4)5approaches 0 and remains positive as xapproaches 4 from the right. This

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