Download Exam 1 - FAU Math

NAME:
Calculus with Analytic Geometry I
Exam 1, Friday, August 30, 2011
SOLUTIONS
1
Solutions to the in class part
1. The graph of a function f is given.
(a) State the value of f (1).
(b) Estimate the value of f (−1).
(c) For what values of x is f (x) = 1?
(d) Estimate the value of x such that f (x) = 0.
(e) State the domain and range of f .
(f) On what interval is f increasing.
Solution.
(a) f (1) = 3.
(b) f (−1) ≈ −1/3 (any value in the range −0.4 to −0.2 is acceptable.)
(c) For x = 1 and x = 3.
(d) f (−1) ≈ −2/3 (any value in the range −0.75 to −0.6 is acceptable.)
(e) The domain is (−2, 4). The range is (−1, 2].
(f) On the interval (−2, 1].
f (a + h) − f (a)
2. If f (x) = x3 , evaluate
. Your final expression should be as simple as possible. That is,
h
anything that can be canceled or simplified in any way, should be canceled and simplified.
2
SOLUTIONS TO THE TAKE HOME PART
2
Solution.
f (a + h) − f (a)
h
=
=
=
(a + h)3 − a3
a3 + 3a2 h + 3ah2 + h3 − a3
6 a3 + 3a2 h + 3ah2 + h3 − 6 a3
=
=
h
h
h
3a2 h + 3ah2 + h3
(3a2 + 3ah + h2 )h
(3a2 + 3ah + h2 ) 6 h
=
=
h
h
6h
3a2 + 3ah + h2 .
2x3 − 5
.
x2 + x − 6
Solution. The only problem this function has is that the denominator can be 0. The quadratic equation
x2 + x − 6 = 0 has the roots 2, −3. The domain is
3. Find the domain of the function f (x) =
(−∞, −3) ∪ (−3, 2) ∪ (2, ∞).
2
Solutions to the take home part
1. Explain what it means that two functions are equal. That is, complete the following (long) sentence. If f
and g are functions, then f = g if and only if
Solution. : If f and g are functions, then f = g if and only if they have the same domain and f (x) = g(x)
for each x in the domain.
Based on your explanation, decide if the following functions f, g are equal or different. In each case, justify
your answer.
(a) f (x) = x2 +
1
1
, g(t) = t2 +
.
1−x
1−t
Equal; the domains are the same and for each point of the domain they assume the same
Solution.
value.
x2 − 1
(b) f (x) =
, g(x) = x − 1.
x+1
Solution. Not really equal. The domain of the first one is (−∞, 1) ∪ (1, ∞), of the second one
(−∞, ∞).
sin u
, g(x) = tan x.
(c) f (u) =
cos u
Solution. Equal. The tangent is by definition the sine over the cosine.
(d) f (x) = eln x , g(x) = x.
Solution. Not equal. The domain of the first one is (0, ∞), of the second one (−∞, ∞).
(e) f (x) = ln(ex ), g(x) = x.
Solution. Equal.
2. Give examples of functions having the following properties:
(a) A function f of domain (−∞, ∞) such that f (s + t) = f (s) + f (t).
Solution. For example, f (x) = 0 for all x. More interesting, f (x) = x.
(b) A function f of domain (−∞, ∞) such that f (s + t) is NOT always equal to f (s) + f (t).
Solution. Almost any function will work. For example, f (x) = x2 .
(c) A function f of domain (−∞, ∞) such that f (s + t) = f (s)f (t).
Solution. Once again f (x) = 0 for all x is an example. More interesting, f (x) = 2x .
2
SOLUTIONS TO THE TAKE HOME PART
3
3. (a) What is a one-to-one function? How can you tell that a function is one-to-one by looking at its graph?
Solution. A one-to-one function is one that assigns different values to different points in the domain.
precisely: A function f of domain D is one-to-one if (and only if) f (s) 6= f (t) whenever s, t ∈ D
and s 6= t. A completely equivalent, sometimes preferred, definition is: A function f of domain D is
one-to-one if (and only if) f (s) = f (t) implies s = t.
From a graph point of view, a function is one-to-one if and only if every horizontal line intersects the
graph of f at most once.
(b) If f is one to one, how is its inverse function f −1 obtained? How are the domain and range of f −1
related to those of f ?
Solution. The inverse function can be obtained by solving for each y in the range of f , the equation
f (x) = y. This gives x as a function of y; that is, it produces x = f −1 (y). One can then, if one wishes,
switch variables to get y = f −1 (x). The domain of f −1 is the range of f , the range of f −1 is the domain
of f .
(c) How do you obtain the graph of f −1 from the graph of f ?
Solution. By reflecting the graph of f with respect to the line y = x.
4. The sine function y = sin x when restricted to the interval [−π/2, π/2] is one-to-one; its inverse is the arc sine
function, denoted by y = sin−1 x or y = arcsin x.
(a) What is the range and domain of y = arcsin x?
Solution. The range of arcsin is the domain of sin (i.e., the domain to which sin was restricted; that
is [−π/2, π/2]. The domain of arcsin is the range of sin, namely [−1, 1].
(b) Simplify: tan(arcsin x). That is, find an expression that does not involve any trigonometric functions.
Solution. I give two solutions. Solution 1., the easy one, by pictures. Think of arcsin x as an angle
θ; that is set th = arcsin x. Then sin θ = x. We have to find tan θ. Draw a right triangle, make on angle
equal to θ and in the simplest way possible give values to a leg and hypotenuse so that sin θ works out
to x. Use the theorem of Pythagoras to find the value of the remaining leg. The simplest way is to say
the
√ hypotenuse equals 1, the leg opposite to the angle is x. Then Pythagoras gives that the other leg is
1 − x2 .
From the picture we can see that tan θ =
√ x
.
1−x2
2
2
Solution 2.,
p no pictures. √Set again θ = arcsin x so sin x = θ. Now sin θ + cos θ = 1 so that
cos2 θ = ± 1 − sin2 θ = ± 1 − x2 . We have to decide if the sign
√ is + or −. The range of arcsin is
[−π/2, π/2]; in that interval cos is non-negative, so that cos θ = + 1 − x2 . thus
tan(arcsin x) = tan θ =
sin θ
x
=√
.
cos θ
1 − x2
5. Find the domain and range of the following functions.
(a)f (x)
=
2
x2 − 4
(b)
g(x)
=
p
64 − x4
(c)h(x)
=
ln(x + 6)
x2 − 9
(d) k(x)
=
1
sin x
2
SOLUTIONS TO THE TAKE HOME PART
4
Solution. In every case to find the range it may pay to try for a more or less primitive graph. I just write
out the answers. You can see me in my office for details.
1
(a) The domain is (−∞, −2) ∪ (−2, 2) ∪ (2, ∞). The range is (−∞, − ) ∪ (0, ∞).
2
√ √
(b) The domain is [−2 2, 2 2]. The range is (0, ∞).
(c) The domain is (−6, −3) ∪ (−3, 3) ∪ (3, ∞). The range is (−∞, ∞).
(d) The domain is {x ∈ (−∞, ∞) : x 6= 0, ±π, ±2π, ±3π, . . .}. The range is (−∞, −1] ∪ [1, ∞).
6. The following picture is the graph of a function. The graph consists of the lower half of the circle of radius 1
centered at the origin, and the upper half of the circle of radius 3 centered at the point of coordinates (4, 0).
Express this function (call it f , for example) as a piecewise defined function.
Solution.
f (x) =
7. Consider the function f (x) =
√
2
−
if −1 ≤ x ≤ 1,
p 1−x ,
9 − (x − 4)2 , if 1 < x ≤ 7.
x+1
.
2x + 1
(a) Determine the domain and the range of f .
(b) Determine the inverse function of f .
Solution. The domain is {x ∈ (−∞, ∞) : x 6= −1/2}. The range is a bit harder to determine, but if we
do part (b) it will work itself out on its own. To find the inverse we solve for x
y=
x+1
;
2x + 1
x=
1−y
.
2y − 1
a bit of algebra gives
1−x
. We found the inverse. We also found the range of f ; since the domain
2x − 1
of the inverse is {x ∈ (−∞, ∞) : x 6= 1/2}, that’s the range of f .
Thus, it seems that f −1 (x) =
1
1
and g(x) = , find an expression for the function f ◦ g and determine its domain.
x
16 − x2
Solution. The domain of f ◦ g is the set of all points in the domain of g which g sends to the domain of
f . To be in the domain of g, we must have x 6= 0; then g sends such an x to 1/x; to be in the domain of f
8. If f (x) = √
2
SOLUTIONS TO THE TAKE HOME PART
5
we need 1/x2 < 16. This works out to x > 4 or x < −4. If x > 4 or if x < −4, it is certainly not 0, so the
domain of f ◦ g is (−∞, −4) ∪ (4, ∞). On that domain,
1
f ◦ g(x) = q
1−
9. Two functions f, g are defined by
−1, if x < 0,
f (x) =
x2 , if x ≥ 0.
.
1
x2
g(x) =
3x2 , if x ≤ 1,
3,
if x > 1
Express g ◦ f as a piecewise defined function.
Solution. If we talk things out,it isn’t too hard. If x < 0, then f (x) = −1. Since −1 ≤ 1, g(−1) =
3(−1)2 = 3. Thus g ◦ f (x) = 3 if x < 0. If x > 0, then f (x) = x2 . What g now does depends on whether
x2 ≤ 1 or x2 ≥ 1. Suppose x2 ≤ 1. Then g(x2 ) = 3(x2 )2 = 3x4 . Notice that since x ≥ 0, the condition x2 ≤ 1
is the same as x ≤ 1. So we have g ◦ f (x) = 3x4 if 0 ≤ x ≤ 1. If x2 > 1; equivalently if x > 1, then g(x2 ) = 3.
Thus g ◦ f (x) = 3 if x > 1. Putting it all together,

x < 0,
 3,
3x4 , 0 ≤ x ≤ 1,
g ◦ f (x) =

3,
x > 1.