Transcript B.5.2 - Concavities and the Second Derivative Test

B.5.2 - Concavities and the
Second Derivative Test
Calculus - Santowski
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Lesson Objectives
• 1. Calculate second and third derivatives of
functions
• 2. Define concavity and inflection point
• 3. Test for concavity in a function using the
second derivative
• 4. Perform the second derivative test to determine
the nature of relative extrema
• 5. Apply concepts of concavity, second
derivatives, inflection points to a real world
problem
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Fast Five
• 1. Solve f’’(x) = 0 if f(x) = 3x3 - 4x2 + 5
• 2. Find the x co-ordinates of the extrema of f(x) = 2x - lnx
• 3. Sketch a graph of a function that has an undefined
derivative at x = c which (i) does and (ii) does not change
concavities.
• 4. Find the 4th derivative of f(x) = x4 + 2x3 + 3x2 - 5x + 7
• 5. If the position, as a function of time, of a vehicle is
defined by s(t) = t3 - 2t2 - 7t + 9, find acceleration at t = 2
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(B) New Term – Graphs Showing
Concavity
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(B) New Term – Concave Up
• Concavity is best “defined”
with graphs
• (i) “concave up” means in
simple terms that the “direction
of opening” is upward or the
curve is “cupped upward”
• An alternative way to describe
it is to visualize where you
would draw the tangent lines 
you would have to draw the
tangent lines “underneath” the
curve
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(B) New Term – Concave down
• Concavity is best “defined”
with graphs
• (ii) “concave down”
means in simple terms that the
“direction of opening” is
downward or the curve is
“cupped downward”
• An alternative way to describe
it is to visualize where you
would draw the tangent lines 
you would have to draw the
tangent lines “above” the curve
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(B) New Term – Concavity
• In keeping with the idea of concavity and the drawn
tangent lines, if a curve is concave up and we were to draw
a number of tangent lines and determine their slopes, we
would see that the values of the tangent slopes increases
(become more positive) as our x-value at which we drew
the tangent slopes increase
• This idea of the “increase of the tangent slope is illustrated
on the next slides:
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(B) New Term – Concavity
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(B) New Term – Inflection Point
• An inflection point is the
point on a function where
the function changes its
concavity (see the black
points on the red curve)
• Mathematically, inflection
points are found where
y’’(x) = 0
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
• Inflection points can also
be found where y’’(x) is
undefined
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(B) New Terms - Concavity
and Inflection Points
• Consider the graphs of
the following
functions and
determine:
• (i) y’’(x)
• (ii) where the
inflection points are
• (iii) what their
intervals of concavity
are
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•
•
•
F(x) = (x - 1)4
G(x) = x^(1/5)
H(x) = x^(2/3)
I(x) = 1/x
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(B) New Terms - Concavity
and Inflection Points
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QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
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(C) Functions and Their
Derivatives
• In order to “see” the connection between a graph of a
function and the graph of its derivative, we will use
graphing technology to generate graphs of functions and
simultaneously generate a graph of its derivative
• Then we will connect concepts like max/min,
increase/decrease, concavities on the original function to
what we see on the graph of its derivative
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(C) Functions and Their
Derivatives
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(C) Functions and Their
Derivatives
•
Points to note:
•
(1) the fcn has a minimum at x=2 and
the derivative has an x-intercept at x=2
•
(2) the fcn decreases on (-∞,2) and the
derivative has negative values on (-∞,2)
•
(3) the fcn increases on (2,+∞) and the
derivative has positive values on (2,+∞)
•
(4) the fcn changes from decrease to
increase at the min while the derivative
values change from negative to positive
•
(5) the function is concave up and the
derivative fcn is an increasing fcn
•
(6) the second derivative graph is
positive on the entire domain
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(C) Functions and Their
Derivatives
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(C) Functions and Their
Derivatives
•
•
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•
•
•
•
•
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f(x) has a max. at x = -3.1 and f `(x) has an xintercept at x = -3.1
f(x) has a min. at x = -0.2 and f `(x) has a root at –
0.2
f(x) increases on (-∞, -3.1) & (-0.2,∞) and on the
same intervals, f `(x) has positive values
f(x) decreases on (-3.1, -0.2) and on the same
interval, f `(x) has negative values
At the max (x = -3.1), the fcn changes from being
an increasing fcn to a decreasing fcn è the
derivative changes from positive values to
negative values
At a the min (x = -0.2), the fcn changes from
decreasing to increasing -> the derivative changes
from negative to positive
f(x) is concave down on (-∞, -1.67) while f `(x)
decreases on (-∞, -1.67) and the 2nd derivative is
negative on (-∞, -1.67)
f(x) is concave up on (-1.67, ∞ ) while f `(x)
increases on (-1.67, ∞) and the 2nd derivative is
positive on (-1.67, ∞)
The concavity of f(x) changes from CD to CU at
x = -1.67, while the derivative has a min. at x = 1.67
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(C) Functions and Their
Derivatives - Summary
• If f ``(x) >0, then f(x) is concave up
• If f `(x) < 0, then f(x) is concave down
• If f ``(x) = 0, then f(x) is neither concave nor concave down, but has an
inflection points where the concavity is then changing directions
• The second derivative also gives information about the “extreme
points” or “critical points” or max/mins on the original function:
 If f `(x) = 0 and f ``(x) > 0, then the critical point is a minimum
point (picture y = x2 at x = 0)
 If f `(x) = 0 and f ``(x) < 0, then the critical point is a maximum
point (picture y = -x2 at x = 0)
• These last two points form the basis of the “Second Derivative Test”
which allows us to test for maximum and minimum values
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(D) Examples - Algebraically
•
Find where the curve y = x3 - 3x2 - 9x - 5 is concave up and concave down. Find and
classify all extreme points. Then use this info to sketch the curve.
•
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f(x) = x3 – 3x2 - 9x – 5
f `(x) = 3x2 – 6x - 9 = 3(x2 – 2x – 3) = 3(x – 3)(x + 1)
So f(x) has critical points (or local/global extrema) at x = -1 and x = 3
•
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f ``(x) = 6x – 6 = 6(x – 1)
So at x = 1, f ``(x) = 0 and we have a change of concavity
•
Then f ``(-1) = -12  the curve is concave down, so at x = -1 the fcn has a maximum
point
•
Also f `(3) = +12  the curve is concave up, so at x = 3 the fcn has a minimum point
•
Then f(3) = -33, f(-1) = 0 as the ordered pairs for the function
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(D) In Class Examples
• ex 1. Find and classify all local extrema using FDT of
f(x) = 3x5 - 25x3 + 60x. Sketch the curve
• ex 2. Find and classify all local extrema using SDT of
f(x) = 3x4 - 16x3 + 18x2 + 2. Sketch the curve
• ex 3. Find where the curve y = x3 - 3x2 is concave up and
concave down. Then use this info to sketch the curve
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3
• ex 4. For the function f (x)  x x  3 find (a) intervals
of increase and decrease, (b) local max/min (c) intervals of
concavity, (d) inflection point, (e) sketch the graph
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(D) In Class Examples
• ex 5. For the function f(x) = xex find (a) intervals of
increase and decrease, (b) local max/min (c) intervals of
concavity, (d) inflection point, (e) sketch the graph
• ex 6. For the function f(x) = 2sin(x) + sin2(x), find (a)
intervals of increase and decrease, (b) local max/min (c)
intervals of concavity, (d) inflection point, (e) sketch the
graph
f (x) 
ln
x
• ex 7. For the function
, find (a) intervals of
increase and decrease, (b) local max/min (c) intervals of
concavity, (d) inflection point, (e) sketch the graph

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(I) Internet Links
• We will work on the following problems in class: Graphing Using First
and Second Derivatives from UC Davis
• Visual Calculus - Graphs and Derivatives from UTK
• Calculus I (Math 2413) - Applications of Derivatives - The Shape of a
Graph, Part II Using the Second Derivative - from Paul Dawkins
• http://www.geocities.com/CapeCanaveral/Launchpad/2426/page203.ht
ml
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(J) Homework
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•
•
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Textbook, p307-310,
(i) Graphs: Q27-32
(ii) Algebra: higher derivatives; Q17,21,23
(iii) Algebra: max/min; Q33-44 as needed +
variety
• (iv) Algebra: SDT; Q50,51
• (v) Word Problems: Q69,70,73
• photocopy from Stewart, 1997, Calculus –
Concepts and Contexts, p292, Q1-26
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