These notes are not going to appear on your exam.
• Finding the domain when the rule ( closed form) of the function is given.
1. Find zeros of the denominator.
2. Find the inequality that makes under the square root negative.
3. Draw a number line. Eliminate all points in part a and b from the number line.
4. What is left of the number line is in form of intervals. Write the intervals in the interval form and
put a union in between them.
• Finding the domain and range using your calculator.
Graph the function. All possible x-values is the domain. All possible y-values is the range.
• Some limit and continuity notes:
1. My recommendation: If you are having trouble answering any questions about a piece-wise
defined function, graph the function first.
2. To find limit of polynomials, plug in the value.
3. To find the limits at infinity, divide numerator and denominator by xn where n is the degree
of the denominator. So only terms with largest degree survive the limit and the rest will vanish.
0
4. : If both numerator and denominators are polynomials, then they have common factors. Simplify
0
and plug in the value.
5. To graph a piece-wise defined function, divide your space into vertical strips which each
strip width is the domain of one piece of function. Then graph each piece of function in the
appropriate strip. Last, decide about open or closed circles ( end points).
6. lim− f (x) is the limit from below.
x→a
7. lim+ f (x) is the limit from above.
x→a
8. If f is defined ”around” x = a ( it doesn’t have to be defined at x = a.), then lim− f (x) = lim+ f (x) =
x→a
x→a
L if and only if lim f (x) = L.
x→a
9. Continuity of Polynomial and Rational Functions
(a) A polynomial function y = P (x) is continuous at every value of x.
(b) A rational function R(x) = p(x)/q(x) is continuous at every value of x where q(x) ≠ 0.
10. f is continuous at x = a if and only if
(1)f (a) is defined, (2)lim f (x) exists and (3)lim f (x) = f (a) .
x→a
x→a
11. If lim+ f (x) or lim− f (x) or lim is infinity or negative infinity then the limit does not exists.
x→a
x→a
x→a
12. The intermediate value theorem.
Let f be continuous on [a, b] and M a value between f (a) and f (b), then there exist number r
in [a, b] such that f (r) = M .
13. Existence of zero of a continuous function.
Let f be continuous on [a, b] and f (a) and f (b) have different signs, then f has a zero in [a, b].
14. How to find algebraic expression (rule) for function f ○ g when algebraic expression
for f and g is given.
Write f with big parenthesis whenever you see x. Write g(x) inside those parenthesis.
15. Four-step process for finding f ′ (x) is
(a) Compute f (x + h). Which is writing out f (x) and replacing x by (x + h) everywhere.
(b) Form the difference f (x + h) − f (x)
f (x + h) − f (x)
.
h
f (x + h) − f (x)
(d) Compute the limit f ′ (x) = lim
.
h→0
h
16. Average rate of change for function f over interval [a, a + h] is:
f (a + h) − f (a)
.
h
17. The instantaneous rate of change of function f at a is:
f (a + h) − f (a)
. ( This limit is also called derivative or f ′ (a) at a.)
lim
h→0
h
18. Notice the difference between Average and instantaneous rate of change.
(c) Form the quotient
19. Differentiability and continuity theorem: If f is differentiable at x = a, then f is continuous
at x = a.
20. f is NOT differentiable at x = a if any of the following cases is true:
(a) Graph of f has corners at x = a.
(b) f is not continuous at x = a.
(c) The tangent line is vertical at x = a.
21. f ′ (a) is the slope of tangent line to graph of y = f (x) at (a, f (a)).
22. The equation of Tangent line at x = a to graph of y = f (x) can be found using the slope-point
formula: y − y0 = m(x − x0 ) where m = f ′ (a), x0 = a and y0 = f (a).
23. Tangent line problems. When f (x) and x = a are given.
(a) Find the derivative. f ′ (x).
(b) Plug in x value a in f ′ (x). If f ′ (a) exists, then the slope of the tangent line. m = f ′ (a)
(c) Plug in a in original f (x) to get y-value f (a).
(d) Use m and (a, f (a)) in the point-slope equation.
y − f (a) = m(x − a).
24. Horizontal tangent line if and only if f ′ (a) = 0.
25. To solve a word problem involving a geometrical shape.
(a) Draw the geometrical shape.
(b) Find a constraint. That is, a number or relation that is forced on the geometrical shape. You
need this to solve all independent variables with respect to one.
(c) Find the function that the problem is asking for. Replace the expressions of independent
variable in the function.
(d) What is the independent variable that was asked in the problem? Go back and use that to
do the blue parts above.
26. Profit=Revenue-Cost
• BASIC RULES OF DIFFERENTIATION
1. Rule 1: The derivative of constant
d
(c) = 0
dx
Derivative of a constant function is zero.
2. Rule 2: The Power Rule: If n is any number, then
d n
(x ) = nxn−1
dx
3. Rule 3: Derivative of constant multiple of a function:
d
d
[cf (x)] = c [f (x)] (c, a constant)
dx
dx
The derivative of a constant times a differentiable function is equal to the constant times the
derivative of a function.
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4. Rule 4: The Sum Rule:
d
d
d
[f (x) ± g(x)] =
[f (x)] ±
[g(x)]
dx
dx
dx
The derivative of the sum(difference) of two differentiable function is equal to the sum(difference)
of their derivatives.
5. Rule 5: THE PRODUCT RULES
d
[f (x)g(x)] = f (x)g ′ (x) + f ′ (x)g(x)
dx
The derivative of product of two functions is Derivative of first times the second plus first times
the derivative of second.
6. Rule 6: THE QUOTIENT RULE
g(x)f ′ (x) − f (x)g ′ (x)
d f (x)
[
]=
dx g(x)
(g(x))2
ow d high minus high d low over the square we go.
7. Rule 7: THE CHAIN RULE
If h(x) = g[f (x)], then
d
g[f (x)] = g ′ [f (x)]f ′ (x)
h′ (x) =
dx
Derivative of composition is the derivative of outside function evaluated at the inside function
times the derivative of inside function.
8. Rule 8: THE GENERAL POWER RULE: This is a less general result of Rule 7.
If the function f is differentiable and h(x) = [f (x)]n , n a real number, then
d
h′ (x) =
[f (x)]n = n[f (x)]n−1 f ′ (x)
dx
• AVERAGE COST FUNCTION:
Suppose C(x) is a total cost function. Then the AVERAGE COST FUNCTION, denoted by
C(x) =
C(x)
x
′
The derivative C (x) of the average cost function, called the MARGINAL AVERAGE COST
FUNCTION, measures the rate of change of the average cost function with respect to the number of
units produced.
• MARGINAL means derivative of whatever follows.
• ELASTICITY OF DEMAND
It is the negative of
P ercentage rate of change of f
percentage rate of change of p
100f (p)
f ′ (p)
100p′
p
=
If f is a differentiable demand function defined by x = f (p), then the ELASTICITY OF DEMAND
at price p is given by
E(p) =
−pf ′ (p)
f (p)
Note: Revenue is R(p) = px = pf (p)
R′ (p) = f (p) + xf ′ (p) = f (p)[1 − E(p)]
• THE DIFFERENTIAL
Let y = f (x) define a differentiable function of x. Then,
1. The differential dx of the independent variable x is dx = △x.
2. The differential dy of the dependent variable y is dy = f ′ (x) △ x = f ′ (x)dx
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3. Note to approximate: △x = xfinal − xinitial and plug in the xinitial or the x value that can be
easily calculated in f ′ (x). Calculate dy
yfinal ≃ yinitial + dy
• Implicit derivation When a function is not given but a relation is given.
d
of both side of the equation.
dx
dy
2. Multiply
next to terms that derivative with respect to y.
dx
dy
3. Solve for
dx
1. Take
• To do related rate problems:
1. Draw a picture if geometrical.
2. Find a relationship between two variable.
3. take the derivative with respect to t.
dx
dy
4. Make sure to put
and
next to appropriate parts.
dt
dt
5. Find or recognize the values for variables and rates of changes that are given replace in the above
and solve for rate that problem is asking.
• A function is increasing/decreasing on an interval (a, b) if for every two numbers x1 and x2 in (a, b),
(f (x1 ) < f (x2 ))/(f (x1 ) > f (x2 )), respectively, whenever x1 < x2 .
• Theorem: Detection of increasing/decreasing for a ” differentiable” function on (a, b)
1. If f ′ (x) > 0 for every value of x in an interval (a, b), then f is increasing on (a, b).
2. If f ′ (x) < 0 for every value of x in an interval (a, b), then f is decreasing on (a, b).
3. If f ′ (x) = 0 for every value of x in an interval (a, b), then f is constant on (a, b).
• Determining the intervals where a function is increasing or decreasing
1. Find all values of x where f ′ (x) = 0 or f or f ′ are discontinuous. Mark them on a line interval
So it will break the number line into intervals.
2. Select a test number c in each interval found and determine the sign of f ′ (c) in that interval.
If f ′ (c) > 0 then f is increasing on the interval.
If f ′ (c) < 0 then f is decreasing on that interval.
• A function f has a relative maximum/ minimum, at x = c if there exists an open interval
(a, b) containing c such that (f (x) ≤ f (c))/ (f (x) ≥ f (c)), respectively, for all x in (a, b).
• A CRITICAL number of a function f is any number x in the domain of f such that
f ′ (x) = 0 or f ′ (x) does not exists.
• Detection of relative Minimum/Maximum of a differentiable in ”enough places” function.
1. Find f ′ (x). Set equal to zero, f ′ (x) = 0 or is discontinuous. Those numbers that are
inside an interval in the domain are called the critical numbers.
2. Either use First derivative test: Draw a number line and find the sign of the derivative
on different intervals.
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(a) If f ′ (x) changes sign from + to − as we move across a critical number c, then f
has a relative maximum at x = c.
(b) If f ′ (x) changes sign from − to + as we move across a critical number c, then f
has a relative minimum at x = c.
(c) If f ′ (x) does not changes as we move across a critical number c, then f does not
have a relative extremum at x = c.
3. Or use second derivative test: Take the second derivative and evaluate at each critical
number c ( for c such that f ′ (c) = 0).
(a) If f ′′ (c) < 0, then f has a relative Maximum at c. (/.)
(b) If f ′′ (c) > 0, then f has a relative Minimum at c. (,.)
(c) If f ′′ (c) = 0 or does not exist, then test is inconclusive.
• Inflection Point: A point on the graph of a continuous function f where the tangent line exists and
where concavity changes.
Examples: The following are examples of inflection points.
c
c
• Finding inflection points of a twice differentiable in ”most places” function.
1. Compute f ′′ (x).
2. Determine the numbers in the domain of f for which f ′′ (x) = 0 or f ′′ does not exists.
3. Determine the sign of f ′′ (x) to the left and right of each number found in previous step. If there
is a change in sign of f ′′ (x) as we move across x = c, then (c, f (c)) is an inflection point of f .
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c