ap12_calculus_bc

AP

®

CALCULUS BC

2012 SCORING GUIDELINES

Question 4

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x

1 1.1 1.2 1.3 1.4



f

x

8 10 12 13 14.5

The function f is twice differentiable for 0x > with

()

115f =

and

()

1 20.f

′′

=

Values of

,

f

′

the derivative of

f

, are given for selected values of x in the table above.

(a) Write an equation for the line tangent to the graph of f at

1.x = Use this line to approximate

()

1.4 .f

(b) Use a midpoint Riemann sum with two subintervals of equal length and values from the table to

approximate

()

1.4

1

.

f

xdx

′



Use the approximation for

()

1.4

1

f

xdx

′



to estimate the value of

()

1.4 .f

Show

the computations that lead to your answer.

(c) Use Euler’s method, starting at

1x = with two steps of equal size, to approximate

()

1.4 .f

Show the

computations that lead to your answer.

(d) Write the second-degree Taylor polynomial for f about

1.x = Use the Taylor polynomial to approximate

()

1.4 .f

(a)

()

115,f =

()

18f

′

=

An equation for the tangent line is

()

15 8 1 .yx=+ −

() ( )

15 8 1.41.4 1 18.2f ≈+ −=

2 :

{

1 : tangent line

1 : approximation

(b)

() ( )( ) ( )( )

1.4

1

0.2 10 0.2 13 4.6dxfx ≈+=

′



() () ()

1.4

1

1.4 1ff fdxx

′

=+



()

15 4.6 9116.4 .f ≈+ =

3 :

1 : midpoint Riemann sum

1 : Fundamental Theorem of Calculus

1 : answer











(c)

() ()()()

10.21.2 8 16.6ff≈+ =

() ()()

1.4 16.6 0.2 12 19.0f ≈+ =

2 :

{

1 : Euler’s method with two steps

1 : answer

(d)

() ( ) ()

()()

2

20

15 8 1 1

2!

0111851

Tx x x

xx

=+ −+ −

+−−=+

() ()()

2

15 8 1.4 1 10 1.4 1 19..14 8f ≈+ −+ − =

2 :

{

1 : Taylor polynomial

1 : approximation

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AP

®

CALCULUS BC

2012 SCORING COMMENTARY

Visit the College Board on the Web: www.collegeboard.org.

Question 4

Overview

Students were presented with a table of values for

f

′ at selected values of x given that f is a twice-differentiable

function. The values for

()

1

f

and

()

1f ′′ are also given. Part (a) asked students to write an equation for the line

tangent to the graph of f at

1x =

and then use this line to approximate

()

1.4 .f Students should have used the

given values for

()

1

f

and

()

1f ′ to construct an equation equivalent to

() ()

11(1).yf f x=′−+ Students could

then substitute

1.4x = to obtain the desired approximation. Part (b) asked students to use a midpoint Riemann

sum with two subintervals of equal length, based on values in the table, to approximate

()

1.4

1

.f

x

dx′

³

They were

then asked to use this approximation to estimate

()

1.4 .f This estimate is obtained by using the midpoint Riemann

sum in the expression

() ( )

1.4

1

1.

f

xdxf

′

+

³

Part (c) asked students to use Euler’s method, starting at 1x = with

two steps of equal size, to approximate

()

1.4 .f

Part (d) asked for the second-degree Taylor polynomial for f

about

1,x =

which was then used to obtain yet another approximation for

()

1.4 .f Students should have used the

given values for

()

1,f

()

1,f ′ and

()

1f ′′ to write the Taylor polynomial.

Sample: 4A

Score: 9

The student earned all 9 points.

Sample: 4B

Score: 6

The student earned 6 points: 2 points in part (a), 1 point in part (b), 1 point in part (c), and 2 points in part (d). In

parts (a) and (d) the student’s work is correct. In part (b) the student presents a correct midpoint Riemann sum, so

the first point was earned. In part (c) the student correctly presents Euler’s method with two steps, so the first

point was earned. The student’s arithmetic error in the last sum leads to an incorrect value for the approximation.

Sample: 4C

Score: 3

The student earned 3 points: 1 point in part (a), no points in part (b), 2 points in part (c), and no points in part (d).

In part (a) the student presents the correct tangent line and earned the first point. The student makes an algebra

error in subsequent work and did not earn the second point. In part (b) the student’s work is incorrect. In part (c)

the student’s work is correct. In part (d) the student uses incorrect values for

()

1,f

()

1,f

′

and

()

1.f

′′