Working toward Rigor versus Bare-bones justification in Calculus

Todd Ericson

Background Info

► Fort Bend Clements HS ► 25 years at CHS after leaving University of Michigan► 4 years BC Calculus / Multivariable Calculus► 2014 School Statistics: 2650 Total Students 45 Multivariable Calculus Students 110 BC Calculus students 200 AB Calculus students ► 2013: 28 National Merit Finalists► BC Calculus AP Scores from 2011 – 2014 5’s : 316 4’s : 44 3’s : 11 2’s : 2 1’s : 0

Coached the 5A Texas State Championship for Men’s Soccer 2014.

Common Topics involving Justification

► Topics and Outline of Justifications:► Continuity at a point► Differentiability at a point► IVT and MVT (Applied to data sets)► Extrema (Both Relative and Absolute) and Critical values / 1st and 2nd Der. Tests► Concavity/Increasing decreasing Graph behavior including Points of Inflection► Justification of over or under estimates (First for Linear Approx, then Riemann Sums)► Behavior of particle motion (At rest , motion: up,down, left, right)► Error of an alternating Series► Lagrange Error for a Series► Convergence of a series► Justification of L’Hopital’s Rule

Both AB and BC topics are listed below.

References for problems

► Justification WS is 3 page document handed out as you entered.

► All documents will be uploaded to my wikispaces account. Feel free to use or edit as necessary.

► http://rangercalculus.wikispaces.com/

► As we work through problems, I will address certain points and thoughts given in document 2.

► Email for questions: todd.ericson@fortbendisd.com

See attached handout for justification outlines

Sample Problem 1

Continuity

Problem 1

1) Given this piecewise function, justify that the function is continuous at x = 2

2 , 2( )

4 4 , 2x x

f xx x

Continuity

Problem 1 Initial Solution (How can we create a more rigorous solution?

► 1) ) (2) 4) lim ( ) 4

2) (2) lim ( )

2( ) is continuous at x = 2

a fb f xx

c f f xx

f x

Continuity

Problem 1 Solution

► 1) ) (2) 4(2) 4 4 (Indicates we are using the appropriate piecewise function)) lim ( ) 4

2lim ( ) 4

2lim ( ) 4 (Indicates we check the limit from both the left and right side)

2) (2) lim ( )

2

a fb f x

xf x

xf x

xc f f x

x

( ) is continuous at x = 2f x

Sample Problem 2

Differentiability

Problem 2

► 2) Given this piecewise function, justify that the function is not differentiable at x = 2

Differentiability

Problem 2 Solution(How can we create a more rigorous solution)?

► 2)

► Or► The function is not continuous at x = 2 therefore it cannot be differentiable

at x = 2.

( ) (2)lim2

2

The function isn't differentiable at x = 2

f x f DNEx

x

Differentiability

Problem 2 Solution

► 2)

► Or

( ) (2)lim ( )2

2( ) (2)lim 4

22

The function f(x) isn't differentiable at x = 2

f x f DNE orx

xf x fx

x

Since lim ( ) lim ( )

2 2lim ( )

2Because of this, the function ( ) is not continuous at x=2 and

therefore cannot be differentiable at x = 2.

f x f x

x xf x DNE

xf x

Sample Problem 3

Extrema

Problem 3

► 3) Find the absolute maximum and minimum value of

the function in the interval from

Extrema

Problem 3 Solution(How can we create a more rigorous solution)?

► 3) x y

0 1

e

1

sin( )'( ) cos( ) xf x x e

Critical point at x = 2

The absolute maximum is ( , ).2

The absolute minimum is (0,1) and ( ,1).

e

Extrema

Problem 3 Solution

► 3)

x y

0 1

e

1

sin( )'( ) cos( ) xf x x e

'( ) 0 at x = 2

f x

The absolute maximum of f(x) is e and occurs at x = .2

The absolute minimum of f(x) is 1 and occurs at x = 0 and x = .

Sample Problem 4

IVT/MVT - Overestimate

Problem 44) Given this set of data is taken from a function v(t) and assuming it is continuous over the interval [0,10] and is twice differentiable over the interval (0,10)

T=0 hours T=1 hour T=2 hours T=4 hours T=6 hours T=10 hours

Vel=50mph Vel=60mph Vel=30mph Vel=38mph Vel=50mph Vel=70mph

a) Find where the acceleration must be equal to 4 mile per hour2 and justify.

b) Find the minimum number of times the velocity was equal to 35mph and justify.

c)Approximate the total distance travelled over the 6 hour time frame starting at t = 4 using a trapezoidal Riemann sum with 2 subintervals.

d) Assuming that the acceleration from 4 to 10 hours is strictly increasing. State whether the approximation is an over or under estimate and why.

IVT/MVT - Overestimate

Problem 4 Solution► a) Given that the function v(t) is continuous over the interval [0,10] and differentiable over the interval (0,10) and since and there must exist at least one c value between hours 2 and 4 such that by the Mean value theorem.

►  b) Given the function v(t) is continuous over the interval [0,10] and since v(1)=60 and v(2) = 30 and since v(2)=30 and v(4) = 38 there must exist at least one value of c between hour 1 and hour 2 and at least one value between hour 2 and hour 4 so that v(c)=35 at least twice by the Intermediate Value theorem.

► c)

► d) Since the function v(t) is concave up and above the x-axis (because the derivative of velocity is increasing) . The top side of the trapezoid will lie above the curve and therefore the approximation will be an over estimate.

10

4

1 1( ) (2)(38 50) (4)(50 70) 328 miles2 2

v t dt

2(4) (2)( ) 44 2

v va c mph

2'( ) 4v c mph

( ) '( )a t v t

Sample Problem 5

Taylor Series

Problem 5

► 5) Given the functions

► a)Find the second degree Taylor Polynomial P2(x) centered at zero for

► b) Approximate the value of using a second degree Taylor Polynomial centered at 0.

► c) Find the maximum error of the approximation for if we used 2 terms of the Taylor series to approximate the value.

3

( ) sin

( )

g x x

h x x

1sin2e

( )( ) g xf x e

312e

Taylor Series

Problem 5 Solution

sin

2 sin sin

2

2

2

5 ) (0) 1

'( ) cos , '(0) 1

' '( ) cos sin , ' ' (0) 1

( ) 12

1 1 1 5) 12 2 8 8

) Since the Taylor Series is an alternating series, with terms decreasing in magnitude and w

x

x x

a f

f x x e f

f x x e x e f

xP x x

b P

c

6 9 3 13

hose values are approaching zero as x , the maximum error of a two term approximationis the absolute value of the third term.

( ): ( ) 1 ..... ...2! 3! ( 1)!

12

nx x xTaylor Series f x xn

Third term

6

1 1The two term Taylor series will have a maximum error of 2! 128 128

Additional Time - Additional Problem

Additional Problem

2014 Problem 3

Additional Problem

2014 Problem 3