Math 115 Spring 2014Written Homework 1

Due Monday, February 3

Instructions: Write complete solutions on separate paper (not spiral bound). If multiplepieces of paper are used, they must be stapled with your name and lecture written on eachpage. In general, answers with no supporting work will not receive full credit. Please reviewthe Course Information document for more complete instructions.

1. In lecture, we quickly reviewed function notation and how to determine different thingsabout the function from its explicit definition. You should also be able to answer questionsabout a function from its graph.The graphs of two functions, f and g, are given below.

Note: These can be answered without showing any work.

(a) Estimate the values of f(0), g(1), and g(f(−4)).Solution: For any function F (x), the graph of F is the collection of points (a, F (a)) where ais in the domain of F . On the graph of f , we see the point (0, 3). Hence f(0) = 3. Similarly,the point (1, 1) is on the graph of g. Hence, g(1) = 1. Finally, for the composition g(f(−4)),we see the point (−4,−2) is on the graph of f , so g(f(−4)) = g(−2). The point (−2, 1) ison the graph of g, so g(−2) = 1 and g(f(−4)) = 1(b) Estimate for what values of x is f(x) ≥ g(x)?Solution: Graphically, f(x) ≥ g(x) is represented when the y coordinates of f are greaterthan y coordinates of g (i.e. the graph of f is “above” the graph g). We see that there aretwo points of intersection, approximately (−2, 1) and (2, 2). In between these two points wesee that f(x) ≥ g(x), so f(x) ≥ g(x) for x in the interval [−2, 2].(c) Estimate for what values of x is f(x) = 2.Solution: Here, the requirement f(x) = 2 requires that we look for the points on the graphof f where the y-coordinate is 2. There are approximately two points on the graph of f thatsatisfy this; (−1.5, 2) and (2, 2). Hence, the solutions to the equation f(x) = 2 are x = −1.5and x = 2.(d) On what interval is f decreasing?Solution: Notice that the y-coordinate for the graph of f decreases after it obtains it max-imum height at the point (0, 3). Hence, the values of f(x) are decreasing on the interval(0, 4].

(e) State the domain and range of f using set notation.Solution: The domain of f is represented by the set of the values on the horizontal axis atwhich there is a point (x, f(x) on the curve. Here, we see that there is a point on the graphcorresponding to all the x values in the interval [−4, 4]. Hence, in set notation the domainof f must be {x ∈ R | − 4 ≤ x ≤ 4}.

Equivalently, the range of f is represented by the set of the values on the vertical axisat which there is a point (x, f(x)) on the curve. We see that there is a point on the graphcorresponding to all the y values in the interval [−2, 3]. Hence, the range of f is

{f(x) ∈ R | − 2 ≤ f(x) ≤ 3}.

(f) State the domain and range of g using interval notation.Solution: Using the same approach as in part (e). The domain of g is [−4, 3] and the rangeof g is [−1/2, 4].2. Find the domain of each function. Express your answer in both set notation and intervalnotation.

(a) f(x) :=x− 3√

x(x2 + 3x + 2)

Since f is a composition of functions including the√x we have the x ≥ 0. We also must

have that the denominator of f is never zero.

x2 + 3x + 2 = 0

(x + 1)(x + 2) = 0

Thus, we have that x 6= −1 or x 6= −2. However, we already have that x ≥ 0, so thesetwo values are not in our domain. Furthermore, since

√x is in the denominator, we must

eliminate the value of x for which√x = 0, and we must exclude x = 0 from our domain.

This leaves, in interval notation(0,∞).

In set notation we have{x ∈ R|x > 0}

(b) g(u) :=√

2− u−√

2 + uThe argument of each square root must be non-negative. This means that 2 − u ≥ 0 and2+u ≥ 0. Simplifying the inequalities gives u ≤ 2 and u ≥ −2 respectively. Since both needto be satisfied we have −2 ≤ u ≤ 2. (Note: this can be thought of as (−∞, 2)

⋂(−2,∞)).

The answer in interval notation is[−2, 2].

The answer in set notation is{u ∈ R| − 2 ≤ u ≤ 2}.

3. The difference quotient for a given function f(x) at the point x = a is defined as

f(a + h)− f(a)

h

For the function f(x) :=x + 1

2x + 3, fully simplify the difference quotient.

Solution:

f(a + h)− f(a)

h=

a+h+12(a+h)+3

− a+12a+3

h

=

2a2+3a+2ah+3h+2a+3−(2a2+2ah+3a+2a+2h+3)(2a+2h+3)(2a+3)

h

=3h− 2h

h(4a2 + 6a + 4ah + 6h + 6a + 9)

=1

4a2 + 4ah + 12a + 6h + 9

4. Determine whether each curve is the graph of a function of x and explain how you know.If it is the graph of a function, state the domain and range of the function using intervalnotation.

Figure 1: part (a) Figure 2: part (b)

(a) Solution: The given curve passes the vertical line test at every point. The domainof the function is [−2, 2] and the range is [−1, 2].(b) Solution: The given curve fails the vertical line test. The vertical line x = c intersectsthe curve at more than just a single point for every integer c.5. Let h(x) :=

√9− x2.

(a) State the domain and range of h using set notation.Solution: Even powered roots require the argument of the root to be non-negative. Thus,we need 9− x2 ≥ 0. This gives

9− x2 ≥ 0

x2 ≤ 9

−3 ≤ x ≤ 3,

and the domain of h is {x ∈ R| − 3 ≤ x ≤ 3}.For the range of h, we need to understand the behavior of the root function. The smallestvalue h can be is zero when x = 3 or x = −3. Since x2 is always non-negative, the term −x2

is less than or equal to zero always. Thus 9 − x2 decreases as x increases, and 9 − x2 willtake its largest value when x = 0. Since

√9− 02 =

√9 = 3. The range of h is

{h(x) ∈ R|0 ≤ h(x) ≤ 3}.

(b) Sketch the curve y = h(x) on the Cartesian plane. (Hint: Algebraic manipulation ofthe equation y = h(x) should revel a familiar curve. Then, make sure your graph correspondsto your answer in part (a).)Solution: To graph h, we sketch the points determined by the equation y =

√9− x2 in

xy-space that also satisfy the domain and range of h. Start with√

9− x2. Squaring bothsides of the equation yields y2 = 9−x2. Manipulation yields x2 +y2 = 9, which we recognizeas the equation of the circle of radius 3 centered at the origin. Thus, we sketch the “top”half of this circle to correspond with the domain and range we found in part (a).

6. Sketch the graph of each function.(a) F (x) := −2(x + 2)Solution: To graph F , we sketch the points determined by the equation y = F (x) in xy-space. With some algebra, we should recognize the equation y = −2(x + 2) represents aline in R2. Distributing the −2, we get y = −2x − 4. This line is now in slope interceptform y = mx + b. The slope of the line is m = −2 and the value b = −4 implies that they-intercept of the line occurs at the point (0,−4). Using this information, results in thefollowing graph.

(b) H(t) :=

{t + 1 if t < −1−2t if t ≥ −1

Note: This is referred to as a piece-wise defined function. It is a function that is defined by

different formulas in different parts of its domainSolution: First, we need to understand that H is a piecewise defined function. H is definedas t + 1 on the piece of its domain t ∈ (−∞,−1). To graph h on this piece, we sketch thepoints determined by the equation y = H(t) in ty-space. You should recognize this as apiece of the line y = t + 1. The slope of the line is m = 1 and the value b = 1 implies thatthey y-intercept of the line y = t+ 1 occurs at the point (0, 1). However, due to the domainrestriction, knowing the y-intercept is not as useful. Instead, we can plug in a value for tthat is in our domain. For instance, H(−2) = −2 + 1 = −1, so the point (−2,−1) is onthis graph. Using the above information and restricting the domain results in the followinggraph. The above graph only corresponds to “half” of the graph of the entire function.

Another piece of H is defined as −2t on the domain [−1,∞). Again, this is part of the liney = −2t. Here the slope is −2 and the y-intercept is (0, 0). Sketching the line and restrictingthe domain results in the following graph.

Combining the two graphs, we get the graph of H.

7. Find a function whose graph is the line segment joining the points (4,−10) and (−8, 8).Solution: The slope of the line segment is defined by

m =−10− 8

4− (−8)=−18

12=−3

2.

Using point slope form, we can find an equation of the line in slope intercept form.

y − 8 = −3

2(x− (−8))

y − 8 = −3

2x− 12

y = −3

2x− 4.

Thus, this line segment lies on the graph of the equation y = −32x− 4, and lies on the graph

of the function L(x) := −32x− 4. However, we are not done yet, as the graph of L(x) is an

infinitely long line. We want just the line segment. To achieve this, we restrict the domainof L. The function whose graph corresponds to just the line segment is f(x) := −3

2x − 4

where −8 ≤ x ≤ 4.8. Let f(x) :=

√x and g(x) := 3

√1− x. Recall that f ◦ g denotes the composition of f

and g which is defined as (f ◦ g)(x) = f(g(x)). For the following functions, (i) show thedomain of the composition by shading the appropriate region of the real number line and(ii) fully simplify the expression.(a) f ◦ g

Solution: (f ◦ g)(x) =√

3√

1− x =[(1− x)1/3

]1/2= (1− x)

13× 1

2 = (1− x)16 or 6√

1− x

Note the the domain of the inside function g is all real numbers. Since the compositionis an even root, we require the argument 1 − x to be non-negative. This requires x ≤ 1.Drawing this interval on the number line:

(b) g ◦ fSolution: (g ◦ f)(x) = 3

√1−√x. The expression is fully simplified.

Note the the domain of the inside function f is x ≥ 0 and the domain of g is all realnumbers. Thus the domain of the composition is just x ≥ 0. Drawing this interval on thenumber line:

(c) f ◦ fSolution: (f ◦ f)(x) =

√√x = 4√x.

Since the composition is an even powered root, the domain of the composition is [0,∞).Drawing the interval on the number line:

9. If g(x) := x + 4 and h(x) := x2 + 8x + 24, find a function f such that f ◦ g = hSolution: We want the function f(x) such that f(g(x)) = h(x). That is, we require f(x +4) = x2 + 8x + 24. When f acts on x + 4, we know that it must square the argument x + 4in order for a x2 term to result in simplified composition. Our first guess for f would bef̃(x) := x2. However, this composition, f̃(x + 4), does not result in h(x):

f̃(x + 4) = (x + 4)2 = x2 + 8x + 16 6= x2 + 8x + 24

We see that f̃(x + 4) is off by the constant 8. Thus, the function

f(x) := x2 + 8

should result in the composition we want. Check:

f(g(x)) = f(x + 4) = (x + 4)2 + 8 = x2 + 8x + 16 + 8 = x2 + 8x + 24 = h(x).