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Mathematics

MATH 1271

Scot Adams

Fall

Description

Priyanka Ketkar
Exam Analysis: Calculus I, Final
Test Breakdown
The main topics covered in the Calculus I final exam were differentiation, limits, function
analysis, functions, continuity, and integration. The most frequently covered topics are
differentiation, function analysis, and integration. Because the final exam covers a greater range
of topics, question types are varied.
The final exam is a total of 21 questions.
Questions 1-15 are multiple choice. There is not a particular topic of question that is too
prevalent in this section, but overall, these questions are straightforward and quick. They are
usually a direct reference to a rule or property and less focused on analysis.
Questions 16-21 are free response and may contain multiple parts. Once again, there is not a
particular topic that is the most common, but the questions in this section require more time and
problem-solving skills. Oftentimes, these questions require clever manipulation of functions and
analysis of the meaning behind certain answers.
Test Statistics
Topics Fall 2005 Fall 2007 Total
Differentiation 4 7 11
Limits 4 3 7
Function Analysis 5 6 11
Functions 1 1
Continuity 1 1
Integration 8 7 15 Topic 1:Differentiation
1. Knowledge Summary
Derivative: slope of a tangent to a curve; instantaneous rate of change; use tangent
formula, but there are different ways to express derivatives
o dy/dx: derivative of y with respect to x
o y’: first derivative of y; also expressed as f’(x)
o df/dx: derivative of a function with respect to x
o d/dx f(x): derivative of function f(x).
Derivatives of polynomials are always one degree less than the original function.
When comparing graphs of functions and their derivatives, maximums and minimums
occur where the slope equals the tangent.
Rules
Derivative of a Constant function: f’(c) = 0
Power Rule: if f(x) = x , then f’(x) = nx – 1
o If more than one term in a function, do this to each individual term.
Constant Multiple Rule: f’(cu) = cf’(x)
Sum and Difference Rule: f’(u ± v) = f’(u) ± f’(v)
Product Rule: f’(uv) = uf’(v) + vf’(u)
o For three functions: f”(uvw) = uvw’ + uv’w + u’vw
Quotient Rule: f’(u/v) = (vf’(u) – uf’(v))/v
Chain Rule: if y = f(g(x)), then y’ = f’(g(x)) * g’(x); The derivative of the outside times
the derivative of the inside. o If there are three functions when the main function is decomposed, do y’ of the
outside * y’ of the second outside * y’ of the inside
Derivatives of Exponential and Logarithmic Functions
L'Hôpital's rule: if a limit is indeterminant, take the derivative of the numerator and the
denominator separately and then put them back together. (Use only for the limit
definition, not as a substitution for the quotient rule). Only works if it is an indeterminant.
x x u u
dy/dx e = e dy/dx e = e * du/dx
x lna^x xlna
a = e = e …. Use this to change terms so you can use exponential derivative rules
dy/dx a = a * du/dx * lna
dy/dx lnu = 1/u *(du/dx)
dy/dx log b = 1/(ulnb) * du/dx
Derivatives of Trigonometric and Inverse Trigonometric Functions
f’sinx = cosx
f’cosx = –sinx
f’tanx = sec x
f’cscx = –(cscx)(cotx)
f’secx = (secx)(tanx)
f’cotx = –csc x
-1 2 1/2
f’sin (u) = 1/(1 – u ) (du/dx)
f’cos (u) = –1/(1 – u ) (du/dx)
f’tan (u) = 1/(1 + u ) (du/dx)
-1 2
f’cot (u) = –1/(1 + u ) (du/dx) -1 2 1/2
f’sec (u) = 1/(|u|(u – 1) ) (du/dx)
f’csc (u) = –1/(|u|(u – 1) ) (du/dx)
Implicit vs. Explicit Differentiation
Explicit Function: in terms of a single variable; y equals some expression with only x and
2 2x
constants; ex: y = 2x + 3x, y = cos(╥ )
o For derivatives, just do dy/dx = and just use previous rules to manipulate x
Implicit Function: the x’s and y’s are together; may or may not be able to use algebra to
make it explicit; ex: y = sin(xy), x + 2xy + y = 14
o To take derivative, take derivative of all the terms on both sides using previous
rules
o If taking derivative of y term, multiply that term by dy/dx; for x terms, multiply
by dx/dx (this cancels because it equals 1). Basically take derivative of everything
with respect to x.
o Use algebra to solve for dy/dx
Differentiability
A function is differentiable at a certain point if the limit of its derivative exists at that
point. Thus all properties (sum, difference, product, quotient, and exponent) of limits
apply to test differentiability.
Functions are not differentiable at
o Corners: usually in piecewise functions; if multiple tangents are possible,
derivative does not exist.
o Cusp (an extreme, curvy corner) o Vertical Tangent: if the slope is going to be vertical, or undefined, derivative does
not exist
o Any type of discontinuity in the function.
*Derivatives exist at endpoints because the limit of the function at that point exists.
Not all continuous functions necessarily have continuous derivatives, but all continuous
derivatives come from continuous functions.
Newton’s Method: The main concept is that when you make a guess as to where a root
of a function is, the x intercept of the tangent line to that curve should be a better
approximation than the initial guess.
o By repeating this process, the approximation gets very close to the root.
o Equation: x = n n-1– (f(xn-1f’(x n-1for the nth approximation
2. Summary of Questions to be Asked
These questions are very common throughout the whole exam. Some questions explicitly
ask to take the derivative of a function, while others implicitly ask to do it for either an
approximation. The questions in the free response section will often have more tedious or
longer calculations than those of section one.
3. Related Past Test Questions
Fall 2007 Final Question 1
Answer: (c)
Here, we must implicitly take the derivative of the given equation. [ ]
( ) ( )
Fall 2007 Final Question 6
Answer: (c) y – 10 = 9(x – 2)
Here, we must take the derivative of the function, evaluate it at the given point to find the
slope there, and use the point and slope to get an equation of a tangent line.
( )
( ) ( )
( )
Fall 2007 Final Question 8
Answer: (b)
( )
Here, we must use the quotient rule to take the derivative.
* + ( )( ) ( )( )
( )
( ) ( )
Fall 2007 Final Question 9 Answer: (d) 2.025
Estimating using linearizations is the same thing as finding the tangent line of a function
and using the plugging a certain value of x into the tangent line equation instead of into
the function. The theory is that if the x value of the point we are estimating is close
enough to the point of tangency, the tangent line will give a relatively good estimation of
the function.
( ) √
( )
( √ )
Since 8.3, the x value for which we are estimating, is very close to 8, we will use x = 8 as
the point of tangency.
( )
(√ ) ( ) ( )
( ) ( ) ( )( )
( ) ( )
( ) ( )
Now we can plug in 8.3 for x.
( ) ( )
Fall 2007 Final Question 16
Part A: Answer:
Here, we must implicitly take the derivative twice and substitute as needed. ( )( ) ( ) ( )
( )
Part B: Answer:
To express y” in terms of only y, we could solve for x and do a traditional substitution.
However, since the only x term we need to get rid of is the x term, we can solve the
2 2 2
original equation for x to get x = 9y + 9. Now, we can plug that into y” and simplify.
( )
Fall 2007 Final Question 19
Part A: Answer: 2
Here, we must refer to the Newton’s Method formula : x = xn n-1– (f(xn-1f’(xn-1 to solve
this problem. Since we are looking for x 3 we must do 2 attempts of the formula.
( ) ( )
( )
( ) ( )
( )
( ) ( )
Fall 2005 Final Question 1
xy xy
Answer: (a) y’ = (ye – 2x)/(2y – xe ) Here, we must use implicit differentiation to find the derivative.
( )
( )
Fall 2005 Final Question 9
Answer: (c) 1.975
Using linear approximation is the same as finding the tangent line at the nearest whole
number x value and then plugging in 7.7 for x into the tangent line.
( ) √
( )
( √ )
( )
(√ ) ( ) ( )
Now we can get the equation of the tangent line, which is
( ) ( ) ( )
( ) ( )
( ) ( )
Fall 2005 Final Question 16
2
Answer: 3x Although we can easily take the derivative of the given function, we are being asked to
use the basic definition of a derivative, which is the limit of a secant slope as h
approaches 0.
( ) ( ) ( )
Fall 2005 Final Question 19
Answer: y = x
To find the tangent line, we need to have the slope, which is the derivative at a certain
point, and a point. At x = 1, f(x) = 1, so our point is (1,1).
Because we have a function of x raised to another function of x, we need to manipulate
the function to take the derivative.
( )
( )
( )
( ) ( )
( ) ( ( ) ) ( ( ) )
( ) ( ( ) ( ) ) ( )
Now, we can make our tangent line equation
( ) Topic 2: Limits
1. Knowledge Summary
The limit of a function is the value the function approaches as x approaches a certain
value.
The expression ( ) refers to the limit as x approaches a from the right.
The expression ( ) refers to the limit as x approaches a from the left.
The expression The expression ( ) refers to the limit as x approaches a from both
sides.
A horizontal asymptote of a function is also the limit of that function to either + or – ∞.
Properties
1) Sum and difference: The limit of a sum/difference of two functions is the
sum/difference of the limits of the separate functions; ( ( ) ( ))
( ) ( )
2) Product or quotient: The limit of a product/quotient of two functions is the
product/quotient of the limits of the separate functions (as long as the
( ) ( )
denominator is not 0 for quotients); or ( ( )
( ) ( )
( ) ) ( ) ( ) 3) Exponents: The limit of a rational power of a function is the power of the limit
of the function (the exponent should not be 0 and the limit should be a real
number); ( ) ( ( )) or ( ) ( )
TRIGONOMETRIC LIMITS
o sin(x) = 0
x→0
o cos(x) = 1
x→0
o tan(x) = 0
x→0
o csc(x) = DNE
x→0
sec(x) = 1
x→0
o
x→0 cot(x) = DNE
o sin(ax) = 1
x→0 ax
1 – cos(ax) = 0
x→0
ax
INDETERMINATES
An indeterminate occurs when two laws of mathematics conflict when evaluating a limit
∞
(for example – 0/0, ∞/∞, ∞+∞, ∞-∞, 1 , etc.). Basically, it means “do more work”.
A finite indeterminate occurs when x approaches a finite number. There are 2 options
for this case:
1. Factor the function, cancel any like factors, and reevaluate the limit.
2. Graph the function and visually inspect what value the function approaches from
the left and right. An infinite indeterminate occurs when x approaches an infinite number. There is one
option for this case: divide the numerator and denominator by x lower of the two degrees of the top and
bottom polynomials
. Then reevaluate the limit.
A harmonic function is 1/x in which x approaches ∞. This function ends up
equaling 0.
*Note: do not unnecessarily cancel out terms and factors in the function unless there is an
indeterminate while evaluating the limit for the first time.
L'Hôpital's Rule says that . That is, you can take the derivative of the
top and bottom separately and then retry the limit.
o This only works if the indeterminate is 0/0 or ∞/∞. If it is something else, make it
a fraction so that the indeterminate becomes 0/0 or ∞/∞.
You cannot keep taking the derivative just to simplify the expression. There must be an
indeterminate of 0/0 in the previous step in order to take the derivative again.
2. Summary of Questions to be Asked
Limits are not as common on the final exam, but when they do show up, they almost
always require the use of L'Hôpital's Rule. The questions in the free response sections
will require more tedious calculations and function manipulation.
3. Related Past Test Questions
Fall 2007 Fall Question 10
Answer: (c) -5/2 By just plugging in 0 for x, we get an indeterminate of 0/0. We can get past this by using
L'Hôpital's Rule.
√ √ √ √
√ √
√ √
Fall 2007 Fall Question 21a
Answer:
By just plugging in 1 for x, we get an indeterminate of ∞ - ∞. We want to be able to use
L'Hôpital's Rule, but first, we must make the expression into a single fraction.
( ) (( ) )
( ) ( ) ( )
( )
( ) ( )
Fall 2007 Fall Question 21b
Answer: e2
By just plugging in -∞, we get an indeterminate. Here, we must manipulate the function
so that it is not a function of x raised to another function of x. Then we can use
L'Hôpital's Rule. ( )
( )
(( ) ) ( )
Now that the expression is rearranged, we can evaluate e raised to the limit of the
exponent, as is allowed by the exponent property of limits. For now, we will just look at
the limit portion and raise e to the evaluated limit at the end.
( )
( ) ( )
( )
-2
Let’s not forget to raise e to the evaluated limit. The answer is e .
Fall 2005 Fall Question 4
Answer: (c) 1
By just plugging in 0 for t, we get an indeterminate of 0/0. So we can use L'Hôpital's
Rule to get past this.
( ( )) ( ( ))( ( )) Fall 2005 Fall Question 11
Answer: (a) 0
By just plugging in 0, we get an indeterminate. In order to use L'Hôpital's Rule, we must
first make the limit into a fraction.
Fall 2005 Fall Question 13
Answer: (d) Does not exist
By just plugging in ∞, we will have an indeterminate. We could go through L'Hôpital's
Rule to prove that this limit does not exist, but just looking at the function itself, we see
that it has a sine function and taking the derivative multiple times will result in either a
sine or cosine function. Because both of these functions oscillate between 1 and -1, we
have no way of telling what its value will be at ∞. Thus, the limit does not exist.
Fall 2005 Fall Question 15
Answer: (d) 48
By just plugging in 0 for x, we will have an indeterminate of 0/0. We want to use
L'Hôpital's Rule, but first, we should use the product property of limits to split it into 3
limits we do not have to deal with the messy product rule when we take the derivative. ( )( )( )
( )( )( )
( )( )( ) ( )( )( )
Topic 3:Function Analysis
1. Knowledge Summary
Extreme Values of Functions
A function is increasing when: x 2 x ,1f(x 2 > f(x1), and f’(x) > 0
A function is decreasing when: x 2 x ,1f(x 2 < f(x1), and f’(x) < 0
Extrema
o Local (relative) max/min: f’(x) = 0, all values around f(x1) are less than 1(x );
many points in one function
o Absolute (global) max/min: usually ∞ or – ∞, but formally, DNE because ∞ is
not a number; only one point in a function
Critical points: points in the domain of f in which f’ is either 0 or does not exist (DNE);
potential max/min points
Absolute max/min on an interval
o Open interval (a,b): absolute max/min may occur at an interior critical point
o Closed interval [a,b]: absolute max/min will occur at an interior critical point or
an endpoint.
o *make sure when finding critical points that they are included in the interval and
know whether the endpoints are inclusive or noninclusive Connecting f’ and f” with the Graph of f
Concave up: y’ is increasing on that open interval; f” > 0; graph sits above its tangent
line; “holds water”
Concave down: y’ is decreasing on that open interval; f” < 0; graph hangs below tangent
line; “dumps out water”
*Near the point of tangency, the tangent line represents the curve (about the same y
values)
st
1 Derivative Test (finds max, min, increasing and decreasing)
o Find first derivative and find its critical points (set equal to 0 or find where DNE).
o Do number line analysis with the critical points as roots
*label beside it what function’s number line analysis you are doing.
o If a region is positive, that means derivative is positive curve is increasing
o If a region is negative, derivative is negative curve is decreasing
o A critical point that its increasing to the left and decreasing to the right is a max
o A critical point that is decreasing to the left and increasing to the right is a min
o If signs are same on both sides, it is not a max or min, but when writing where it
is increasing or decreasing, do not just ignore that point since the function is still
neither increasing nor decreasing there.
2 Derivative Test (finds inflection points, concave up, and concave down)
o Find the second derivative; zeros and DNE are possible inflection points (PIPs)
o Do number line analysis
o Positive region concave up
o Negative region concave down o If there is a sign change, the PIP in between is an inflection point
With all this information and the roots of the original function, it is possible to graph it
without a calculator.
*If first derivative changes increasing/decreasing but second one does not, there is a cusp
*If first derivative stays same, but second derivative changes, there is a vertical tangent
2. Summary of Questions to be Asked
This is another common topic on the final exam. While the question types may vary, the
processes used to answer the questions are much the same. Most questions involve
analyzing the first and second derivatives of a function to find extrema and inflection
points. These questions often look for regions in which the graph is increasing/decreasing
or concave up/down. While the questions in the free response sections are not necessarily
more difficult than those of the multiple choice section, they may require obtaining more
information of a graph.
3. Related Past Test Questions
Fall 2007 Final Question 3
Answer: (a) (-∞,-3/2]
Here, we are given the first derivative, so we must set it equal to 0, solve for x, and do a
number line analysis. A rational function is equal to 0 when its numerator equals 0. By
setting 2x + 3 = 0, we get x = -3/2.
A function also has critical points when a derivative does not exist, such as when its
denominator equals 0. By setting the denominator equal to 0, we get x = 1. For all x values less than -3/2, f’(x) is negative. For all values of x between -3/2 and 1 as
well as greater than 1, f’(x) is increasing. Since we are looking for the region in which
f(x) decreases, we need the region in which f’(x) is negative, which is (-∞,-3/2].
Fall 2007 Final Question 4
Answer: (c) f(x) has absolute minima at √ =and a local, but not absolute, maximum
at x = 0
We must take the first derivative, set it equal to 0, solve for x, and do a number line
analysis.
( ) ( ) ( √ )( √ )
We get x = 0, x =√ , and x √ . f’(x) is negative for values of x le√s andn
between 0 and√ . f’(x) is positive for values of x b√tweand 0 and greater than
√ . With this information, we can say that there are local minim√ aand x =√
and a local maximum at x = 0.
Since f(x) is of an even degree, its end behavior will be that it will approach ∞ as x
approaches ∞ and -∞. From this, we know that the maximum at x = 0 is only local, since
the function attains higher values as x approaches extreme values.
Since the function does not approach -∞, we know that at least one of the local minima is
a global minimum. For our two local minima, we can plug the x values into the function
to see which one attains the lowest value. In this ca√e, and√ as x values make
f(x) equal to -9. Thus, both local minima are global minima.
Fall 2007 Final Question 7 Answer: (d)√
Here, we are not directly given any function to work with. We must use key words to
extract functions and derivatives.
From the first sentence, we see the word “rate”, so we know that we have a derivative.
Because we are dealing with side lengths and diagonal lengths, we must establish
equations for those.
√ √
Since the end value we are looking for is the rate at which the diagonal increases, we
must find the derivative of the diagonal length.
√
We do not directly know the rate at which the side length increases, so we must find it
through the area equation and then substitute it.
√
Since we know that the derivative of area is 6 and s = 3 in this case, we can plug them in
to find the rate at which the diagonal increases at this point. √ ( ) √ √
Fall 2007 Final Question 11
Answer: (e) f(x) has a local maximum at x = 1, a local minimum at x = 2, and an
inflection point at x = 3/2.
Here, we are looking for local extrema as well as inflection points, so we need to conduct
a number line analysis with the first and second derivatives of the function.

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