math 147 exam i practice problemsglahodny/math147/spring2017...2. solve each inequality. express...
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Math 147 Exam I Practice Problems
This review should not be used as your sole source for preparation for the exam. Youshould also re-work all examples given in lecture, all homework problems, all lab assignmentproblems, and all quiz problems.
1. Solve each of the following equations.
(a) |2x− 5| = 3
(b) |x+ 3| = |2x+ 1|
2. Solve each inequality. Express your answer using interval notation.
(a) |5x− 2| < 6
(b) |3x+ 2| ≥ 4
3. Find an equation of the line passing through (−1, 2) and (3,−4). Express your answerin slope-intercept form.
4. Find an equation of the line passing through (5, 2) and parallel to the line 4x+6y = −5.Express your answer in slope-intercept form.
5. Find an equation of the line passing through (4,−1) and perpendicular to the linepassing through (−2, 1) and (1,−2). Express your answer in slope-intercept form.
6. Find an equation of the circle with radius 3 and center (2,−5).
7. Show that the equationx2 + y2 + 2x− 6y + 7 = 0
represents a circle. Find the center and radius of the circle.
8. Find the radian measure of 210◦.
9. Find the degree measure of 5π/4 radians.
10. Find the exact trigonometric ratios for θ = 2π/3.
11. Find all values of α in the interval [0, 2π) such that
2 sin2 α = 1
12. Evaluate each expression.
(a) log6136
(b) log3 108− log3 4
(c) log2 6− log2 3 + log2 4
(d) 3log3 2+log3 5
1
13. Solve each equation for x.
(a) ln(4x− 1) = 5
(b) e5−2x = 3
(c) log x+ log(x− 3) = 1
(d) 34x = 5
14. Find the domain and range of each function:
(a) f(x) =2
3x− 5
(b) g(x) =√
2x− 5
15. Polonium 210 has a half-life of 140 days.
(a) If a sample has a mass of 300 mg, find a formula for the mass that remains aftert days.
(b) When will the mass be reduced to 40% of its original amount? (Express youranswer in terms of the natural logarithm and include the appropriate unit fortime.)
16. If f(x) =√x and g(x) =
√2− x, find the functions f ◦ g and g ◦ f and state their
domains.
17. Show that f(x) =√−1− x is one-to-one and find its inverse function.
18. Sketch the graph of each function by applying appropriate transformations.
(a) y = −(x− 2)2 + 1
(b) y = −2 +1
x+ 1
(c) y = −ex−3 + 4
(d) y =1
3sin(x− π
4
)19. Use a logarithmic transformation to find a linear relationship between appropriate
transformations of x and y if y = 2× 74x.
20. Use a logarithmic transformation to find a linear relationship between appropriatetransformations of x and y if y = 4x5.
2
21. Use the semilog plot below to find a functional relationship between x and y. Expressyour answer in the form y = a · bx.
0 1 2 3 4
x
10-2
10-1
100
101
102
y
22. Use the double log plot below to find a functional relationship between x and y. Expressyour answer in the form y = axb.
100
101
x
102
103
104
105
y
23. Evaluate each of the following limits, if they exist.
(a) limx→2
x2 − 4
x− 2
(b) limx→3−
x
x2 − 2x− 3
(c) limx→0
sin 7x
x
(d) limx→1
√x2 + 1−
√x+ 1
x− 1
(e) limx→−∞
x3 − x2 + 1
1− x2(f) lim
x→∞(e−x sinx)
3
24. Determine the largest interval on which f(x) =ln(1− x)
ln(1 + x)is continuous.
25. Find the value of c which makes the given function continuous on R = (−∞,∞).
f(x) =
{x2 + 1, x ≤ 1x− c, x > 1
26. Consider the function f defined as
f(x) =
x− 4a if x < −2b if x = −2ax2 if x > −2
where a and b are fixed constants.
(a) Find limx→−2−
f(x) and limx→−2+
f(x).
(b) Find the value of a for which limx→−2
f(x) exists.
(c) For the value of a found above, what is limx→−2
f(x)?
(d) For the value of a found above, determine the value of b for which f is continuousat x = −2.
27. Prove that there is a root of the equation
x5 − 2x4 − x− 3 = 0
in the interval (2, 3).
28. Use the bisection method to approximate a root of the equation
x4 + x3 + x− 1 = 0
with maximum error less than 13.
29. Consider the function f(x) =√x+ 2.
(a) Find f ′(x) using the limit definition of the derivative.
(b) Find an equation of the tangent line to the graph of y =√x+ 2 at x = 14.
Express your answer in slope-intercept form.
30. Consider the function f(x) =3
x.
(a) Find f ′(x) using the limit definition of the derivative.
(b) Find an equation of the normal line to the graph of y =3
xat x = 1. Express your
answer in slope-intercept form.
4
Solutions
Solutions may contain errors or typos. If you find an error or typo, please notify me [email protected].
1. Solve each of the following equations.
(a) |2x− 5| = 3
Using properties of the absolute value, |2x− 5| = 3 is equivalent to
2x− 5 = 3 or 2x− 5 = −3
So 2x = 8 or 2x = 2. Thus, x = 4 or x = 1.
(b) |x+ 3| = |2x+ 1|
Using properties of the absolute value, |x+ 3| = |2x+ 1| is equivalent to
x+ 3 = 2x+ 1 or x+ 3 = −2x− 1
So x = 2 or 3x = −4. Thus, x = 2 or x = −4/3.
2. Solve each inequality. Express your answer using interval notation.
(a) |5x− 2| < 6
Using properties of the absolute value, |5x− 2| < 6 is equivalent to
−6 < 5x− 2 < 6
−4 < 5x < 8
−4
5< x <
8
5
Therefore, the solution set is the open interval (−45, 85).
(b) |3x+ 2| ≥ 4
Using properties of the absolute value, |3x+ 2| ≥ 4 is equivalent to
3x+ 2 ≥ 4 or 3x+ 2 ≤ −4
In the first case, 3x ≥ 2, which gives x ≥ 23. In the second case, 3x ≤ −6, which
gives x ≤ −2. So the solution set is
(−∞,−2] ∪ [23,∞)
5
3. Find an equation of the line passing through (−1, 2) and (3,−4). Express your answerin slope-intercept form.
The slope of the line is
m =−4− 2
3− (−1)= −3
2
Using the point-slope form with (−1, 2), we obtain
y − 2 = −3
2(x+ 1)
which can be expressed in slope-intercept form as
y − 2 = −3
2x− 3
2
y = −3
2x+
1
2
4. Find an equation of the line passing through (5, 2) and parallel to the line 4x+6y = −5.Express your answer in slope-intercept form.
The equation of the given line can be rewritten as
y = −2
3x− 5
6
which has slope m = −2/3. Parallel lines have the same slope, so the required line hasslope −2/3 and its equation in point-slope form is
y − 2 = −2
3(x− 5)
Converting to slope-intercept form, we have
y − 2 = −2
3x+
10
3
y = −2
3x+
16
3
6
5. Find an equation of the line passing through (4,−1) and perpendicular to the linepassing through (−2, 1) and (1,−2). Express your answer in slope-intercept form.
The slope of the line through the points (−2, 1) and (1,−2) is
m =−2− 1
1− (−2)= −1
Perpendicular lines have slopes which are negative reciprocals, so the required line hasslope 1 and its equation in point-slope form is
y − (−1) = 1(x− 4)
Converting to slope-intercept form, we have
y = x− 5
6. Find an equation of the circle with radius 3 and center (2,−5).
Using the standard form for the equation of a circle, we obtain
(x− 2)2 + (y + 5)2 = 9
7. Show that the equationx2 + y2 + 2x− 6y + 7 = 0
represents a circle. Find the center and radius of the circle.
First, we group the x and y terms as follows:
(x2 + 2x) + (y2 − 6y) = −7
Then we complete the square in both x and y, adding the appropriate constants toboth sides of the equation:
(x2 + 2x+ 1) + (y2 − 6y + 9) = −7 + 1 + 9
(x+ 1)2 + (y − 3)2 = 3
Thus, the center of the circle is (−1, 3) and the radius is√
3.
7
8. Find the radian measure of 210◦.
To convert from degrees to radians, we multiply by π/180. Therefore,
210◦ = 210( π
180
)=
7π
6radians
9. Find the degree measure of 5π/4 radians.
To convert from radians to degrees, we multiply by 180/π. Therefore,
5π
4radians =
5π
4
(180
π
)= 225◦
10. Find the exact trigonometric ratios for θ = 2π/3.
A reference triangle for θ = 2π/3 is shown below.
√3
1
2
x
y
π3
Therefore, by definition of the trigonometric ratios, we have
sin θ =
√3
2cos θ = −1
2tan θ = −
√3
csc θ =2√3
sec θ = −2 cot θ = − 1√3
11. Find all values of α in the interval [0, 2π) such that 2 sin2 α = 1.
Solving the given equation for sinα, we obtain
sin2 α =1
2
sinα = ± 1√2.
We are interested in triangles whose opposite sides have length 1 (signed positive ornegative) and whose hypotenuse has length
√2. These triangles are produced by the
angles x = π/4, 3π/4, 5π/4, and 7π/4.
8
12. Evaluate each expression.
(a) log6136
Using properties of exponentials, we obtain
log6
1
36= log6(36−1)
= log6[(62)−1]
= log6(6−2) = −2
(b) log3 108− log3 4
Using properties of logarithms, we obtain
log3 108− log3 4 = log3
108
4= log3 27
= log3(33) = 3
(c) log2 6− log2 3 + log2 4
Using properties of logarithms, we obtain
log2 6− log2 3 + log2 4 = log2
(6 · 4
3
)= log2 8
= log2(23) = 3
(d) 3log3 2+log3 5
Using properties of logarithms, we obtain
3log3 2+log3 5 = 3log3 10 = 10
9
13. Solve each equation for x.
(a) ln(4x− 1) = 5
Exponentiating both sides of the equation, we obtain
eln(4x−1) = e5
4x− 1 = e5
4x = e5 + 1
x =1
4(e5 + 1)
(b) e5−2x = 3
Taking the natural logarithm of both sides of the equation, we obtain
ln(e5−2x) = ln 3
5− 2x = ln 3
2x = 5− ln 3
x =1
2(5− ln 3)
(c) log x+ log(x− 3) = 1
Using properties of logarithms, we obtain
log(x2 − 3x) = 1.
Exponentiating both sides of the equation (with base 10), we obtain
x2 − 3x = 10
x2 − 3x− 10 = 0
(x+ 2)(x− 5) = 0
Therefore, x = −2, or x = 5. However, we must reject x = −2 since it does notlie in the domain of log x. Thus, the only solution is x = 5.
(d) 34x = 5
Taking the (base 3) logarithm of both sides of the equation, we obtain
4x = log3 5
Taking the (base 4) logarithm of both sides of the equation, we obtain
x = log4(log3 5)
10
14. Find the domain and range of each function:
(a) f(x) =2
3x− 5
Since division by zero is undefined, f(x) is undefined if 3x − 5 = 0 or x = 5/3.Thus, the domain of f is (
−∞, 5
3
)∪(
5
3,∞)
In order for f(x) to be defined, either 3x − 5 < 0 or 3x − 5 > 0. If 3x − 5 < 0,then
−∞ <2
3x− 5< 0
Similarly, if 3x− 5 > 0, then
0 <2
3x− 5<∞
Therefore, the range of f is (−∞, 0) ∪ (0,∞).
(b) g(x) =√
2x− 5
Since the square root of a negative number is not defined (as a real number), g(x)is defined if and only if
2x− 5 ≥ 0
2x ≥ 5
x ≥ 5
2
Thus, the domain of f is [5
2,∞)
The function g(x) is defined for all x such that 0 ≤ 2x− 5 <∞. Therefore,
0 ≤√
2x− 5 <∞
Thus, the range of g is [0,∞).
11
15. Polonium 210 has a half-life of 140 days.
(a) If a sample has a mass of 300 mg, find a formula for the mass that remains aftert days.
Let y(t) denote the mass (in mg) of Polonium 210 after t ≥ 0 days. Then
y(t) = y0ekt = 300ekt
Since the half-life of this substance is 140 days,
y(140) = 300e140k = 150
Therefore, we obtain
e140k =1
2
140k = ln
(1
2
)k =
ln 1/2
140
Thus, the mass of Polonium 210 remaining after t days is
y(t) = 300e(ln 1/2140 )t mg
Since eln 1/2 = 1/2, an equivalent expression is
y(t) = 300
(1
2
)t/140
mg
(b) When will the mass be reduced to 40% of its original amount? (Express youranswer in terms of the natural logarithm and include the appropriate unit fortime.)
By part (a), the mass of Polonium 210 left after t days is
y(t) = 300e(ln 1/2140 )t mg
To determine the time at which the mass is reduced to 40% of its original amount,we set
e(ln 1/2140 )t = 0.4
ln 1/2
140t = ln 0.4
t =140 ln 0.4
ln 0.5days
12
16. If f(x) =√x and g(x) =
√2− x, find the functions f ◦ g and g ◦ f and state their
domains.
By definition,
(f ◦ g)(x) = f(g(x)) =
√√2− x = 4
√2− x
Since the fourth root of a negative number is not defined (as a real number), (f ◦ g)(x)is defined if and only if 2− x ≥ 0. That is, if x ≤ 2. Therefore, the domain of f ◦ g is(−∞, 2].
Similarly,
(g ◦ f)(x) = g(f(x)) =
√2−√x
For√x to be defined, we must have x ≥ 0. For
√2−√x to be defined, we must have
2 −√x ≥ 0. That is,
√x ≤ 2 or x ≤ 4. Thus, we have 0 ≤ x ≤ 4 and the domain of
g ◦ f is [0, 4].
17. Show that f(x) =√−1− x is one-to-one and find its inverse function.
The graph of f(x) =√−1− x can be obtained from the graph of f(x) =
√x by
reflecting about the y-axis and then shifting one unit to the left.
By the Horizontal Line Test, the function f is one-to-one. Thus, an inverse functionexists.
To find the inverse function, let y =√−1− x. Interchanging x and y, we obtain
x =√−1− y
x2 = −1− yx2 + 1 = −y−x2 − 1 = y
Therefore, the inverse function is f−1(x) = −x2 − 1.
13
18. Sketch the graph of each function by applying appropriate transformations.
(a) y = −(x− 2)2 + 1
We obtain the desired graph by starting with the parabola y = x2, shifting 2 unitsto the right, reflecting about the x-axis, and shifting 1 unit upward.
(b) y = −2 +1
x+ 1
We obtain the desired graph by starting with the hyperbola y =1
x, shifting 1 unit
to the left, and shifting 2 units downward.
14
(c) y = −ex−3 + 4
We obtain the desired graph by starting with the exponential y = ex, shifting 3units to the right, reflecting about the x-axis, and shifting 4 units upward.
(d) y =1
3sin(x− π
4
)We obtain the desired graph by starting with the sine function y = sinx, shiftingπ
4units to the right, and compressing vertically by a factor of 3.
15
19. Use a logarithmic transformation to find a linear relationship between appropriatetransformations of x and y if y = 2× 74x.
Using a logarithmic transformation,
y = 2× 74x
log y = log(2× 74x)
log y = log 2 + log(74x)
log y = log 2 + 4x log 7
The linear relationship islog y = (4 log 7)x+ log 2
20. Use a logarithmic transformation to find a linear relationship between appropriatetransformations of x and y if y = 4x5.
Using a logarithmic transformation,
y = 4x5
log y = log(4x5)
log y = log 4 + log(x5)
log y = log 4 + 5 log x
The linear relationship islog y = 5 log x+ log 4
16
21. Use the semilog plot below to find a functional relationship between x and y. Expressyour answer in the form y = a · bx.
0 1 2 3 4
x
10-2
10-1
100
101
102
y
Two points on the line are
(x1, y1) = (0, 10−1) and (x2, y2) = (4, 10)
This is a semilog plot where the y axis is on a logarithmic scale. Taking a logarithmictransformation of the y coordinate, we have
(x1, Y1) = (0,−1) and (x2, Y2) = (4, 1)
The slope of the line passing through these points is
m =1− (−1)
4− 0=
1
2
Using slope-intercept form, the equation of the line is
Y =1
2x− 1
Therefore, we have
log y =1
2x− 1
y = 10x/2−1
y =1
1010x/2
y =1
10
(√10)x
17
22. Use the double log plot below to find a functional relationship between x and y. Expressyour answer in the form y = axb.
100
101
x
102
103
104
105
y
Two points on the line are (x1, y1) = (1, 400) and (x2, y2) = (10, 40, 000). Taking alogarithmic transformation, we have
(X1, Y1) = (0, log 400) and (X2, Y2) = (1, log 40, 000)
The slope of the line passing through these points is
m =log 40, 000− log 400
1− 0= log
40, 000
400= log 102 = 2
Using slope-intercept form, the equation of the line is
Y = 2X + log 400
Therefore, we have
log y = 2 log x+ log 400
log y = log(400x2)
y = 400x2
18
23. Evaluate each of the following limits, if they exist.
(a) limx→2
x2 − 4
x− 2
Factoring the numerator yields
limx→2
x2 − 4
x− 2= lim
x→2
(x+ 2)(x− 2)
x− 2= lim
x→2(x+ 2) = 4
(b) limx→3−
x
x2 − 2x− 3
Factoring the denominator yields
limx→3−
x
(x+ 1)(x− 3)=
(limx→3−
x
x+ 1
)(limx→3−
1
x− 3
)= −∞
(c) limx→0
sin 7x
x
Make the substitution u = 7x. Then as x→ 0, it follows that u→ 0. Therefore,
limx→0
sin 7x
x= lim
u→0
sinu
u/7= 7 lim
u→0
sinu
u= 7
(d) limx→1
√x2 + 1−
√x+ 1
x− 1
Rationalizing the numerator yields
limx→1
√x2 + 1−
√x+ 1
x− 1
(√x2 + 1 +
√x+ 1√
x2 + 1 +√x+ 1
)= lim
x→1
x2 + 1− (x+ 1)
(x− 1)(√x2 + 1 +
√x+ 1)
= limx→1
x2 − x(x− 1)(
√x2 + 1 +
√x+ 1)
= limx→1
x(x− 1)
(x− 1)(√x2 + 1 +
√x+ 1)
= limx→1
x√x2 + 1 +
√x+ 1
=1
2√
2
19
(e) limx→−∞
x3 − x2 + 1
1− x2
The largest power of x that appears in the denominator is x2. Therefore,
limx→−∞
x3 − x2 + 1
1− x2= lim
x→−∞
x3 − x2 + 1
1− x2
(1/x2
1/x2
)= lim
x→−∞
x− 1 + 1x2
1x2 − 1
=−∞−1
= ∞
(f) limx→∞
(e−x sinx)
For all x ∈ (−∞,∞), we have
−1 ≤ sinx ≤ 1
−e−x ≤ e−x sinx ≤ e−x
limx→∞
(−e−x) ≤ limx→∞
(e−x sinx) ≤ limx→∞
e−x
0 ≤ limx→∞
(e−x sinx) ≤ 0
By the Sandwich Theorem (Squeeze Theorem),
limx→∞
(e−x sinx) = 0
24. Determine the largest interval on which f(x) =ln(1− x)
ln(1 + x)is continuous.
The domain of any logarithmic function is (0,∞). Thus, in order for f(x) to be defined,we must have
1− x > 0 and 1 + x > 0
That is, −1 < x < 1. Moreover, f(x) is discontinuous if
ln(1 + x) = 0
1 + x = 1
x = 0
Therefore, the function f is continuous on (−1, 0) ∪ (0, 1).
20
25. Find the value of c which makes the given function continuous on R = (−∞,∞).
f(x) =
{x2 + 1, x ≤ 1x− c, x > 1
For x < 1 or x > 1, the function is a polynomial and, therefore, is continuous. Theonly possible point of discontinuity is x = 1. The function is defined at x = 1 sincef(1) = 2. In order for the limit of f(x) as x approaches 1 to exist, set
limx→1−
f(x) = limx→1+
f(x)
limx→1−
(x2 + 1) = limx→1+
(x− c)
2 = 1− cc = −1
If c = −1, thenlimx→1
f(x) = 2 = f(1)
Thus, the function is continuous for c = −1.
26. Consider the function f defined as
f(x) =
x− 4a if x < −2b if x = −2ax2 if x > −2
where a and b are fixed constants.
(a) Find limx→−2−
f(x) and limx→−2+
f(x).
Evaluating the left-hand limit,
limx→−2−
f(x) = limx→−2−
(x− 4a) = −2− 4a
Evaluating the right-hand limit,
limx→−2+
f(x) = limx→−2+
ax2 = 4a
21
(b) Find the value of a for which limx→−2
f(x) exists.
In order for limx→−2
f(x) to exist, the left- and right-hand limits in part (a) must be
equal:
4a = −2− 4a
8a = −2
Thus, a = −1/4.
(c) For the value of a found above, what is limx→−2
f(x)?
If a = −1/4, then using either the left- or right-hand limits in part (a), we have
limx→−2
f(x) = 4a = −1
(d) For the value of a found above, determine the value of b for which f is continuousat x = −2.
To ensure that f is continuous at x = −2, we must have
limx→−2
f(x) = f(−2)
−1 = b
Thus, if a = −1/4 and b = −1, then f is continuous at x = −2.
27. Prove that there is a root of the equation
x5 − 2x4 − x− 3 = 0
in the interval (2, 3).
Let us begin by defining the function f(x) = x5− 2x4−x− 3. Since f is a polynomial,it is continuous on R = (−∞,∞). Observe that
f(2) = 32− 32− 2− 3 = −5 < 0 and f(3) = 243− 162− 3− 3 = 75 > 0
Thus, f(2) < 0 < f(3). By the Intermediate Value Theorem, there exists a number cbetween 2 and 3 such that f(c) = 0. That is, the equation x5− 2x4− x− 3 has a rootc in the interval (2, 3).
22
28. Use the bisection method to approximate a root of
x4 + x3 + x− 1 = 0
with maximum error less than 13.
Let us begin by defining the function f(x) = x4 + x3 + x− 1. Since f is a polynomial,it is continuous on R = (−∞,∞). Observe that
f(0) = 0 + 0 + 0− 1 = −1 < 0 and f(1) = 1 + 1 + 1− 1 = 2 > 0
Thus, f(0) < 0 < f(1). By the Intermediate Value Theorem, there exists a number cbetween 0 and 1 such that f(c) = 0. That is, the equation x4 + x3 + x− 1 = 0 has aroot c in the interval (0, 1).
Consider the midpoint of the interval (0, 1):
x =1 + 0
2=
1
2
and compute
f
(1
2
)=
1
16+
1
8+
1
2− 1 = −15
16< 0
Therefore, a root must lie on the interval (12, 1).
Consider the midpoint of the interval (12, 1):
x =12
+ 1
2=
3
4
The distance from the midpoint to either endpoint is 14< 1
3, so the approximation
x = 34
is sufficient. (Note: This equation has a second real root between −2 and −1,so it’s possible to approximate that root instead.)
23
29. Consider the function f(x) =√x+ 2.
(a) Find f ′(x) using the limit definition of the derivative.
Using the limit definition of the derivative, we have
f ′(x) = limh→0
f(x+ h)− f(x)
h
= limh→0
√x+ h+ 2−
√x+ 2
h
= limh→0
√x+ h+ 2−
√x+ 2
h
(√x+ h+ 2 +
√x+ 2√
x+ h+ 2 +√x+ 2
)= lim
h→0
x+ h+ 2− (x+ 2)
h(√x+ h+ 2 +
√x+ 2)
= limh→0
h
h(√x+ h+ 2 +
√x+ 2)
= limh→0
1√x+ h+ 2 +
√x+ 2
=1
2√x+ 2
(b) Find an equation of the tangent line to the graph of y =√x+ 2 at x = 14.
Express your answer in slope-intercept form.
The slope of the tangent line to the graph at x = 14 is given by
f ′(14) =1
2√
14 + 2=
1
8
The y-coordinate corresponding to x = 14 is given by
f(14) =√
14 + 2 = 4
Using point-slope form, the tangent line is
y − 4 =1
8(x− 14)
Converting to slope-intercept form, we obtain
y − 4 =1
8x− 7
4
y =1
8x+
9
4
24
30. Consider the function f(x) =3
x.
(a) Find f ′(x) using the limit definition of the derivative.
Using the limit definition of the derivative, we have
f ′(x) = limh→0
f(x+ h)− f(x)
h
= limh→0
1
h
(3
x+ h− 3
x
)= lim
h→0
3
h
(x
x(x+ h)− x+ h
x(x+ h)
)= lim
h→0
3
h
(− h
x(x+ h)
)= lim
h→0− 3
x(x+ h)
= − 3
x2
(b) Find an equation of the normal line to the graph of y =3
xat x = 1. Express your
answer in slope-intercept form.
The y-coordinate corresponding to x = 1 is
y =3
1= 3
The slope of the tangent line at x = 1 is given by
f ′(1) = −3
1= −3
Therefore, the slope of the normal line is 13. Using point-slope form, the normal
line is
y − 3 =1
3(x− 1)
Converting to slope-intercept form, we obtain
y − 3 =1
3x− 1
3
y =1
3x+
8
3
25