Exam 1, MATH 1410, Summer 2013
(SOLUTIONS)
HUMANS ONLY! Calculators are NOT allowed.
1. (8 pts.) For which of the following is y a function of x? Write “Yes” or “No” following the statement.
A brief explanation or equation is required to defend your work.
√
(a) {(1, 3), (2, 5), (3, 7), (4, 2), (6, 4), (8, 6), (1, 5)}
(b) x2 y = 1.
(c) x2 y = 0.
(d) x2 y − y = 1.
Solution.
√
(a) {(1, 3), (2, 5), (3, 7), (4, 2), (6, 4), (8, 6), (1, 5)} does NOT give a function since the input x =
√
1 has two outputs, 3 and 5.
1
(b) x2 y = 1 DOES give a function since y is always equal to 2 ; this gives a unique output y in
x
terms of x.
(c) x2 y = 0 does NOT give a function since the input x = 0 has an infinite number of outputs y.
1
(d) x2 y − y = 1 DOES give a function since y is always equal to 2
; this gives a unique
x −1
output y in terms of x.
2. (6 pts.) Graph the function f (x) =
2x + 2,
4−x
if x ≤ 1
if x > 1
=⇒ Then compute f (0), f (1), f (2) and f (100).
Solution. The graph of this function is a line of slope 2 rising from the left, reaching the point
(1, 4) and ending there. (Place a closed circle at (1, 4).)
Then the graph drops down to (1, 3) (place an open circle at (1, 3)) and the graph continues as a
line of slope −1, dropping away to the right.
f (0) = 2(0) + 2 = 2 ; f (1) = 2(1) + 2 = 4 , f (2) = 4 − 2 = 2 and f (100) = 4 − 100 = −96 .
3. (9 pts.) Find the domain of the following functions. Put your answer in interval notation.
√
(a) f (x) = 20 − x.
√
(b) g(x) =
20 − x
.
x
√
(c) h(x) =
20 − x
.
x2 − 4
Solutions.
(a) The domain of f (x) =
√
20 − x is (−∞, 20] .
1
√
20 − x
(b) The domain of g(x) =
is (−∞, 0) ∪ (0, 20] since x = 0 would create a denominator
x
of zero and so cannot be part of the domain.
√
20 − x
(c) The domain of h(x) = 2
is (−∞, −2) ∪ (−2, 2) ∪ (0, 20] since both x = −2 and x = 2
x −4
would create a denominator of zero and so cannot be part of the domain.
4. (10 pts.) Suppose f (x) = 10x2 − x + 16. Simplify
(a)
(b)
(c)
(d)
f (a + 2).
f (a + 2b).
f (x + h).
f (x + h) − f (x).
Solution.
(a) f (a + 2) = 10(a + 2)2 − (a + 2) + 16 = 2(a2 + 4a + 4) + 4a + 8 − 6 = 2a2 + 12a + 10 .
(b) f (a+2b) = 10(a+2b)2 −(a+2b)+16 = 2(a2 +4ab+6b2 )+4a+8b−6 = 2a2 + 8ab + 8b2 + 4a + 8b − 6 .
(c) f (x+h) = 10(x+h)2 −(x+h)+16 = 2(x2 +2xh+h2 )+4x+4h−6 = 2x2 + 4xh + 2h2 + 4x + 4h − 6 .
(d) f (x + h) − f (x) = [answer to (c)] − (10x2 − x + 6). The elements of f (x) cancel out (they
always will!) and we get 20xh + 10h2 − h = h(20x + 10h − 1) .
5. (12 pts.)
The graph of y = f (x) is drawn in red above. For each of the graphs, below, (drawn in blue) first
describe the transformation that turns the above graph into the new graph and then express this
transformation algebraically in terms of the original function f (x).
(a)
(d)
(b)
(e)
(c)
(f)
2
Solutions.
(a) The graph is shifted left 1 unit; y = f (x + 1)
(b) The graph is shifted 4 units left then 2 units up; y = f (x + 4) + 2
(c) The graph is reflected across the x-axis; y = −f (x).
(d) The graph is stretched vertically be a factor of two; y = 2f (x)
(e) The graph is reflected across the x-axis and stretched vertically be a factor of two; y = −2f (x)
(f) The graph is stretched horizontally by a factor of 2, then shifted right by 3 and up by 2;
x
y = f ( − 3) + 2.
2
6. (9 pts.) For each of the functions below, compute f (−x) and use your answer to determine if f (x)
is an even function, an odd function or neither.
(a) f (x) = 2x6 + 4x3 + 8x2 + 10.
(b) f (x) = 2x6 + 8x2 + 10.
(c) f (x) =
2x6 + 8x2 + 10
.
x3 + 4x
Solution.
(a) f (x) = 2x6 + 4x3 + 8x2 + 10 so f (−x) = 2x6 − 4x3 + 8x2 + 10 .
Since this is neither equal to f (x) nor equal to f (−x) then f (x) is NEITHER even nor odd.
(b) f (x) = 2x6 + 8x2 + 10 so f (−x) = 2x6 + 8x2 + 10 = f (x) so f (x) is EVEN .
(c) f (x) = f (x) =
2x6 + 8x2 + 10
2x6 + 8x2 + 10
so f (−x) =
= −f (x) so f (x) is ODD .
3
x + 4x
−x3 − 4x
7. (8 pts.) Suppose f is the function f = {(1, 3), (2, 5), (3, 7), (4, 2), (6, 4), (8, 6)} and g is the function
g = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}. Compute:
(a) (g ◦ f )(2)
(b) (g ◦ f )(6)
(c) (f ◦ g)(2)
(d) (f ◦ g)(6)
Solutions.
(a) (g ◦ f )(2) = g(f (2)) = g(5) = 6 .
(b) (g ◦ f )(6) = g(f (6)) = g(4) = 5 .
(c) (f ◦ g)(2) = f (g(2)) = f (3) = 7 .
(d) (f ◦ g)(6) does not exist (DNE) since 6 is not in the domain of g.
8. (16 pts.) Compute, algebraically, the inverse of each of the following functions.
(x − 4)3 + 12
5
10
(b) f (x) =
6x − 8
(a) f (x) =
3
5x
3x + 4
(d) h(x) = x2 + 6x + 9.
(c) h(x) =
Solutions.
(x − 4)3 + 12
(x − 4)3 + 12
, set
. Then swap inputs and
5
5
(a) To compute the inverse of f (x) =
outputs and solve for y.
x=
(y − 4)3 + 12
,
5
multiply both sides by 5
5x = (y − 4)3 + 12,
and then subtract 12 from both sides
5x − 12 = (y − 4)3 ,
and then take the cube root of both sides
√
3
5x − 12 = y − 4,
and finally add 4 to both sides so that
f −1 (x) =
(b) To compute the inverse of f (x) =
and solve for y.
√
3
5x − 12 + 4.
10
10
, set y ==
. Then swap inputs and outputs
6x − 8
6x − 8
x=
10
;
6y − 8
multiply both sides by the denominator so that
x(6y − 8) = 10,
and then multiply out the lefthand side.
6xy − 8x = 10.
Isolate y by adding 8x to both sides:
6xy = 10 + 8x.
Now divide both sides by 6x so that
y=
10 + 8x
.
6x
Oyr answer is
f −1 (x) =
(c) To compute the inverse of h(x) =
solve for y.
10 + 8x
6x
5x
5x
, set y =
. Then swap inputs and outputs and
3x + 4
3x + 4
x=
4
5y
.
3y + 4
Clear denominators by multiplying both sides by 3y + 4:
x(3y + 4) = 5y.
Multiply out the lefthand side:
3xy + 4x = 5y.
Now we need to isolate y so collect the terms involving y on the left:
3xy − 5y = −4x.
Factor y from the left:
y(3x − 5) = −4x
and finish isolating y by dividing both sides by 3x − 5.
y=−
4x
.
3x − 5
Our final answer is
h−1 (x) = −
4x
.
3x − 5
(d) To compute the inverse of h(x) = x2 + 6x + 9, first notice that h(x) is a perfect square and
write
h(x) = (x + 3)2 or y = (x + 3)2 .
Swap variables
x = (y + 3)2 ;
take the square root of both sides:
√
x=y+3
(which forces y ≥ −3 so our original domain must have required x ≥ −3) and now we add 3
to both sides so that
√
h−1 (x) = x + 3 .
9. (12 pts.)
(a) Find the slope of the line through the points (4, 5) and (6, 17).
(b) Find the equation of the line through the points (4, 5) and (6, 17). Put your answer in pointslope form.
(c) Find the average rate of change of the function f (x) = 3(x − 4)2 + 5 as x varies from 4 to 6.
(d) The line y = g(x) has slope 4 and passes through the point (x0 , 100). Find g(x0 + 2).
Solutions.
17 − 5
12
=
= 6.
6−4
2
y−5
(b) 6 =
so y − 5 = 6(x − 4) .
x−4
(c) The average rate of change of the function f (x) = 3(x − 4)2 + 5 as x varies from 4 to 6 is the
slope of the line joining the point (4, f (4)) to (6, f (6)). Since f (4) = 5 and f (6) = 17, this is
just the slope of the line through the points (4, 5) and (6, 17). We found that in problem (a);
the answer is
17 − 5
m=
= 6.
6−4
(a) m =
5
(d) If the line y = (g) has slope 4 and passes through (x0 , 100) then the y-value will rise by 2(4)
as we move two places to the right. So f (x0 + 2) = 100 + 8 = 108 .
10. (8 pts.)
(a) Complete the square on x2 + 6x by filling in the blanks:
x2 + 6x +
= (x +
)2
so
)2 −
x2 + 6x = (x +
(b) Put the quadratic function f (x) = 5x2 + 30x + 50 in standard form a(x − h)2 + k.
Solution.
(a) x2 + 6x + 9 = (x + 3 )2 so x2 + 6x = (x + 3 )2 − 9 .
(b) 5x2 + 30x + 50 = 5(x2 + 6x) + 50. From part (a) we see that x2 + 6x = (x + 3)2 − 9 so
5x2 + 30x + 50 = 5(x2 + 6x) + 50 = 5((x + 3)2 − 9) + 50 = 5(x + 3)2 − 45 + 50 = 5(x + 3)2 + 5 .
11. (8 pts.) The graph of the quadratic function y = 3(x − 4)2 + 5 is a parabola.
(a) Find the vertex of the parabola.
(b) Explain carefully (in order) the three steps required to transform the graph of y = x2 into
the graph of y = 3(x − 4)2 + 5.
(c) Explain carefully (in order) the three steps required to transform the graph of y = 3(x−4)2 +5
back into the graph of y = x2 .
Solutions. In part (b) we see that the transformations are (in this order)
(a) shift right by 4, then
(b) stretch vertically by a factor of 3 and then
(c) shift up by 5.
If we do this to the origin (0, 0) then we get the vertex (4, 5) .
To answer part (c), just reverse the steps (in reverse order!) (1) Shift down by 5 and then (2) shrink
vertically by a factor of 3 and then (3) shift left by 4.
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