Lesson 4-QR
Quiz 1 Review
Objectives
• Prepare for the quiz on sections 4-1 thru 4-4
Vocabulary
• Critical number – a value of x such that f’(x) = 0 or
f’(x) does not exist
• Existence Theorem – a theorem that guarantees that
there exists a number with a certain property, but it
doesn’t tell us how to find it.
• Extreme values – maximum or minimum functional
values (y-values)
• Indeterminate Form – (0/0 or ∞/∞) a form that a value
cannot be assigned to without more work
• Inflection point – a point (x,y) on the curve where the
concavity changes
Theorems
Extreme Value Theorem: If f is continuous on the closed
interval [a,b], then f attains an absolute maximum value
f(c) and a absolute minimum value f(d) at some numbers
c and d in [a,b].
Fermat’s Theorem: If f has a local maximum or minimum
at c, and if f’(c) exists, then f’(c) = 0. Or rephrased: If f
has a local maximum or minimum at c, then c is a
critical number of f.
[Note: this theorem is not biconditional (its converse is
not necessarily true), just because f’(c) = 0, doesn’t
mean that there is a local max or min at c!!
Example y = x³]
Closed Interval Method:
To find the absolute maximum and minimum values
of a continuous function f on a closed interval [a,b]:
1. Find the values of f at the critical numbers of f in
(a,b) (the open interval)
2. Find the values of f at the endpoints of the
interval, f(a) and f(b)
3. The largest value from steps 1 and 2 is the
absolute maximum value; the smallest of theses
values is the absolute minimum value.
Theorems
Mean Value Theorem:
Let f be a function that is
a) continuous on the closed interval [a,b]
b) differentiable on the open interval (a,b)
then there is a number c in (a,b) such that
instantaneous rate of change = average rate of change
f(b) – f(a)
f’(c) = --------------b–a
mtangent
=
msecant
Rolle’s Theorem:
Let f be a function that is
a) continuous on the closed interval [a,b]
b) differentiable on the open interval (a,b)
c) f(a) = f(b)
then there is a number c in (a,b) such that f’(c) = 0
Review of 1st and 2nd Derivatives
• Function
– Extrema are y-values of the function!
• First Derivative – f’(x)
– Slope of the function
– f’(x) = 0 at “critical” values of x
• Possible locations of relative extrema
• Relative extrema can also occur at endpoints on closed
intervals [a,b]
• Second Derivative – f’’(x)
– Concavity of the function
– f’’(x) = 0 at possible points of inflection
• IPs are places where there is a change in concavity
1st and 2nd Derivative Tests
• First derivative test uses the change in signs of the
slopes [f’(x)] just before and after a critical value to
determine if it is a relative min or max
Relative Max
f’ > 0 for x < c
f’ < 0 for x > c
Slope
+ 0 hill
Relative Min
f’ < 0 for x < c
x=c
Slope
- 0 +
valley
f’ > 0 for x > c
x=c
• Second derivative test uses a functions concavity at the
critical value to determine if it is a relative min or max
Relative Max
Relative Min
x=c
x=c
f’’(x) < 0
Concave down
f’’(x) > 0
Concave up
Intervals - Table Notation
f(x) = x4 – 4x3
= x3(x – 4)
f’(x) = 4x3 – 12x2 = 4x2(x – 3)
f’’(x) = 12x2 – 24x = 12x (x – 2)
(-∞,0)
0
0,2)
2
(2,3)
3
(3,4)
4
(4,∞)
f(x)
+
0
-
-16
-
-27
-
0
+
f’(x) - slope
-
0
-
-16
-
0
+
64
+
f’’(x) - concavity
+
0
-
0
+
36
+
96
+
Intervals
Notes:
IP
y=0
IP
min
y=0
L’Hosptial’s Rule
• L’Hospital’s Rule applies to indeterminate quotients
in the form of 0/0 or ∞/∞
f(x)
f’(x)
Lim -------- = Lim --------g(x)
g’(x)
(can be applied several times)
• Other indeterminate forms exist and can be solved
for, but are beyond the scope of this course
Extrema Problem
f(x) = x³ + 4x² -12
f’(x) = 3x² + 8x = x (3x + 8)
f’(x) = 0 at x = 0 and x = -8/3
f’’(x) = 6x + 8
f’’(x) = 0 at x = -4/3
Intervals
(-∞,-8/3)
f(x)
-8/3
(-8/3,-4/3)
-68/27
Remember: if we have a
closed interval, then we
have to evaluate the end
points!
-4/3
(-4/3,0)
-196/27
0
(0,∞)
-12
f’(x) - slope
+
0
-
-
-
0
-
f’’(x) - concavity
-
-
-
0
+
+
+
Notes:
Rel
max
IP
Rel
min
Mean Value Theorem
f(x) = 6x² - 3x + 8
on [0,3]
Continuous on [0,3]
polynomial
Differentiable on (0,3) polynomial
f(3) = 53
f(0) = 8
53 – 8
45
f’(c) = ------------ = ------- = 15
3–0
3
M secant
f’(x) = 12x – 3 = 15
M tangent
MVT applies
12x = 18
x = 3/2
Mean Value and Rolle’s Theorems
Continuous on [a,b]
problems where f(x) is undefined
division by zero not allowed
negative numbers in an even root
Differentiable on (a,b)
problems where f’(x) is undefined
division by zero not allowed
negative numbers in an even root