Final Review Partial Solutions
Problem 5: Using the double angle identity for sin, we have
sin(2x) sin x = cos x
⇒ (2 sin x cos x) sin x = cos x
⇒ 2 sin2 x cos x = cos x
⇒ 2 sin2 x cos x − cos x = 0
⇒ cos x(2 sin2 x − 1) = 0
π
which is true if cos x = 0 or 2 sin2 x − 1 = 0. cos x = 0 implies x = (2n + 1) , while solving 2 sin2 x − 1 = 0 yields
2
√
2
π
π
π
sin x = ±
, which occurs when x = (2n + 1) . The solutions are x = (2n + 1) or x = (2n + 1) , for n = 0, 1, 2, . . . .
2
4
2
4
Problem 6: Use the fundamental identity 1 + tan2 x = sec2 x, we have
3
sec x
2
3
⇒ sec2 x − 1 = sec x
2
3
⇒ sec2 x − sec x − 1 = 0.
2
tan2 x =
The last equation is a quadratic equation with sec x instead of just x. Regarding sec x as the variable, we can use the
quadratic formula to yield
q 3 2
3
±
− 4(1)(−1)
2
2
,
sec x =
2
which after simplifying shows that sec x = 2 or sec x = − 21 . To determine x, we solve the corresponding equations
1
sec x = 2 and sec x = − 12 . Since sec x =
, the second equation says that cos x = −2, which cannot happen for any x
cos x
(since −1 < cos x < 1). However, the first equation sec x = 2, is equivalent to cos x = 12 , which occurs when x = π3 , 5π
3 ,
etc. This can be concisely written as x = { π3 (6n + 1), π3 (6n + 5)}, n = 0, 1, 2, . . .
4π
4π
1
Problem 7: To determine the exact value of sin−1 (cos
), we notice that cos
= − . Thus, we must determine the
3
3
2
1
7π 11π
π
π
−1
value of sin (− ). The angles for which this is satisfied are
,
or { (12n + 7), (12n + 11)}, n = 0, 1, 2, . . .
2
6
6
6
6
1 + sin x
Problem 15: To solve
= 2, we can multiply by sin x to yield the equation 1 + sin x = 2 sin x, which can be
sin x
reduced to the equation sin x = 1. This occurs for angles in degrees 90 ± 360n, n = 0, 1, 2, . . .
1
11π
11π
Problem 16: One might think that tan−1 (tan
)=
, since tan−1 and tan are inverse functions. However, recall
3
3
π π
that the range of tan−1 x is the interval (− , ), thus the solution should be in this interval. The angle in this interval,
2 2
11π
5π
which is equivalent to
is
.
3
3
Problem 17: To determine the value of csc(tan−1 5), let y = tan−1 5 which is equivalent to tan y = 5. That is, for the
angle y the tangent value is equal to 5. One method to proceed is to setup a right triangle with angle y. Since tan y = 5,
we have that side of the triangle opposite of y is of length 5, while the adjacent side is of length 1. Notice now, that
determining the value of csc(tan−1 5) is equivalent to determining csc y, which is defined as the quotient of the hypotenuse
and the opposite side of the triangle. The length of the hypotenuse h can be determined by the Pythagorean theorem,
√
√
hyp. √
= 26. Thus, csc(tan−1 5) = 26.
h = 52 + 1. Since the opposite side has length 1, we have csc y =
opp.
Problem 18: To determine an algebraic expression for cot(cos−1 x), we can apply the same idea as for 17). That is,
define y = cos−1 x, which is equivalent to cos y = x. This last equation corresponds to a right triangle with angle y,
adjacent
p side x and hypotenuse 1. Again, the opposite side length a can be determined by the Pythagorean theorem:
x
a = ± 1 − x2 . Now, since cot y is equal to the adjacent length divided by the opposite length, we have cot y = √
.
± 1 − x2
x
So, cot(cos−1 x) = √
.
± 1 − x2
Problem 21: The identity
− cos3 x − cos2 x sin x + cos x + sin x
= sin x + cos x
1 − cos2 x
can be verified by manipulating the left hand side. To wit, we have
− cos3 x − cos2 x sin x + cos x + sin x
− cos2 x(cos x + sin x) + (cos x + sin x)
=
1 − cos2 x
1 − cos2 x
(cos x + sin x)(− cos2 x + 1)
=
1 − cos2 x
= cos x + sin x.
Problem 22: Let f (x) = 3x x2 − 4 · 3x . To find the y-intercept, we set x = 0: f (0) = 30 02 − 4 · 30 = 0 − 4 = −4.. So, the
y-intercept is (0, −4).
To find the x-intercept, we set y = 0 and solve for x. That is, we solve
0 = 3x x2 − 4 · 3x .
Factoring out the term 3x , we have
0 = 3x (x2 − 4),
2
and the product of two numbers can only equal 0 if one of the products is 0. That is, we have either
3x = 0
or x2 − 4 = 0.
By the properties of exponential functions, we know that the exponential function 3x can never equal 0. Thus, we only
have the case x2 − 4 = 0, which has solutions x = ±2. So, the x-intercepts are (±2, 0).
Problem 23: To determine the value b such that the point (−2, e) is on the graph of f (x) = bx , we set x = −2 and
y = e, then solve for b. That is, we solve the equation
e = b−2 .
Taking natural logs of both sides of this equation yields
ln(e) = ln(b−2 )
or
1 = −2 ln(b),
using the properties of logarithms. Thus, we have
1
ln(b) = − ,
2
which upon taking exponentials of both sides yields
1
b = e− 2 .
x
x
1
= e− 2 .
Plugging b back into the function f yields f (x) = e− 2
Problem 26: To solve the equation
81x − 30(9x ) = −81,
notice that 81 = 92 , so the equation can be rewritten as
(92 )x − 30(9x ) + 81 = 0.
Slightly rewriting this result yields the equation
(9x )2 − 30(9x ) + 81 = 0,
which is a quadratic equation with variable 9x rather than just x. Thus, we can solve for the unknown 9x using the
quadratic formula. This, gives the solutions
p
30 ± (−30)2 − 4(1)(81)
9x =
,
2
which after simplification yields 9x = 28 or 9x = 2. To determine x we can take log9 of the last two equations to yield
x = log9 28
or x = log9 2.
Problem 28: To solve the logarithmic equation
log4 x = 1 − log4 (x − 3)
3
we can rearrange to yield
log4 x + log4 (x − 3) = 1.
Using the rules of logarithms (namely, that loga x + loga y = loga (xy)), we have the equation
log4 (x2 − 3x) = 1.
Taking exponentials (with base 4) of both sides yields
2
4log4 (x
−3x)
= 41 ,
which simplifies to
x2 − 3x = 4.
This last quadratic equation can be solved to yield x = −1 or x = 4.
Problem 32: The range of the exponential functions ax is the interval (0, ∞) for any base a > 0. Thus the range of
3x is (0, ∞). Making the exponent negative (i.e., 3−x ) reflects the graph about the y-axis, hence the range does not
change. However, the range of −3−x is (−∞, 0) since the negative out front reflects the graph about the x-axis. Adding
−5 to this yields −5 − 3−x , where the −5 shifts the graph of −3−x down by 5 units. All together, we have the range of
f (x) = −5 − 3−x is the interval (−∞, −5).
Problem 33: To solve the inequality
which can be rewritten as
and further rewritten as
x
1
< 16, we first solve the equality
4
x
1
= 16,
4
x
1
= 42
4
x −2
1
1
=
.
4
4
Taking log1/4 of both sides yields
1
1
x log1/4 ( ) = −2 log1/4 ( )
4
4
which has solution x = −2. To determine the inequality that x satisfies, we can check the original inequality for values
of x < −2 and for values of x > −2. After picking a value in each interval, the final inequality should be x > −2.
Problem 37: Recall that ln x has domain (0, ∞), so ln x√ is defined if x > 0. Thus, ln(4x2 − 5) is defined when 4x2 − 5 > 0.
Solving this inequality shows that x must satisfy ±x > 25 . Multiplying ln(4x2 − 5) by 3 does not affect
the domain,!since
√ !
√
5
5
2
this stretches the graph vertically. So, the domain of f (x) = 3 ln(4x − 5) is the interval −∞, −
∪
,∞ .
2
2
4
Problem 47: Verify the identity
cosh(2x) − cosh2 x = sinh x.
The definitions of the hyperbolic cosine and sine are
cosh x =
ex + e−x
2
sinh x =
ex − e−x
.
2
Plugging the definition of cosh x into the left hand side of the identity yields
2
cosh(2x) − cosh x =
=
=
=
x
2
e + e−x
e2x + e−2x
−
2
2
1 2x 1 −2x
1 2x 1 x −x 1 −x x 1 −2x
e + e
−( e + e e + e e + e
)
2
2
4
4
4
4
1 2x 1 −2x 1 2x 1 −2x 1
e + e
− e − e
−
2
2
4
4
2
e2x + e−2x
1
− .
4
2
Now, plugging the definition of sinh x into the right hand side of the original identity yields
2
ex − e−x
2
1 2x 1 x −x 1 −x x 1 −2x
= e − e e − e e + e
4
4
4
4
e2x + e−2x
1
=
− .
4
2
sinh2 x =
Since the left and right hand sides yield the same result, the identity holds.
Problem 53: To solve the inequality log5 (x − 4) > 2, we first solve the equality log5 (x − 4) = 2. Taking exponentials
with base 5 of both sides yields
5log5 (x−4) = 52
which implies
x − 4 = 52 .
The solution is then x = 29. To determine the solution of the inequality, we can simply check the inequality log5 (x−4) > 2
for a value in the interval x > 29 and for a value in x < 29. I chose the two values 129 and 9. Then
log5 (129 − 4) = log5 (125) = 3 > 2,
since 53 = 125. Finally,
log5 (9 − 4) = log5 (5) = 1 ≯ 2,
1
since 5 = 5.
5