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Mean Value Theorem Worksheet

Mean Value Theorem Worksheet
  • Page 1
 1.  
We verify the "Mean Value theorem" for a function f (x):
a.
If it is continuous on [a, b]
b.
If it is differentiable on (a, b)
c.
If it is continuous on (a , b)
d.
Both A and B


Solution:

If f(x) is continuous on [a, b] and is differentiable on (a, b), then only it follows the hypotheses of mean value theorem for which the mean value theorem can be verified.


Correct answer : (4)
 2.  
State whether the function f (x) = 3x2 - 2 on [2, 3] satisfies the mean value theorem.
a.
no
b.
yes


Solution:

f (x) = 3x2 - 2, x [2, 3]
[Given function.]

The given function is continuous on [2, 3] and differentiable in (2, 3).

So, there exists c (2, 3) such that f ′ (c) = f(3) - f(2)3 - 2.
[By mean value Theorem.]

f ′(c) = 15.
[f ′(c) = 6c. ]

6c = 15 c = 5 / 2 (2, 3).

Hence, the given function satisfies the mean value theorem.


Correct answer : (2)
 3.  
Which of the following functions satisfies the mean value theorem?
a.
f(x) = x2 - 6 , x [7, 8]
b.
f (x) = log x, x [-7, 8]
c.
f (x) = [x], x [-7, 7]
d.
f (x) = |x|, x [-8, 8]


Solution:

f (x) = x2 - 6, the function f (x) is continuous on [7, 8] and differentiable on (7, 8)
[Consider the choice A.]

f ′ (c ) = f (8) - f (7)8 - 7 = 15
[Try to get c (7, 8) such that f ′ (c ) = f (b) - f (a)b - a]

2c = 15
[f ′ (x) = 2x]

c = 15 / 2 [7, 8]
[There exists c (7, 8) such that f ′ (c ) = f (b) - f (a)b - a]

So, the function in the choices A satisfies the mean value theorem

f (x) = log x, x [- 7, 8] is not continuous in [- 7, 8] and hence does not satisfy the mean value theorem
[Consider the choice B.]

f (x) = [x], x [- 7, 7] is not continuous in [- 7, 7] and hence does not satisfy the mean value thoerem
[Consider the choice C.]

f (x) = |x|, x [- 8, 8] is continuous in [- 8, 8] but not differentiable in (- 8, 8) and hence does not satisfy the mean value theorem .
[Consider the choice D.]


Correct answer : (3)
 4.  
Determine the point on the parabola f (x) = (x - 2)2, at which the tangent is parallel to the chord joining the points (2, 0) & (3, 1).
a.
(5 2, 1 4)
b.
(3 2, 1 4)
c.
(3 2, 9 4)
d.
(3 2, - 3 2)


Solution:

Slope of the chord joining (2, 0) & (3, 1) = 1-0 / 3-2 = 1

Slope of the tangent to the curve at any point (x, f(x)) is f ′ (x) = 2(x - 2)

f ′ (c) = 1
[By Mean Value Theorem.]

2(c - 2) = 1

c = 5 / 2 (2, 3)

f (c) = (c - 2)2 = (5 / 2 - 2)2 = 1 / 4

Hence, the point where the tangent to the parabola is parallel to the given chord is (c, f (c)) = (5 / 2, 1 / 4)


Correct answer : (1)
 5.  
Find the point on the parabola y = (x + 3)2, at which the tangent is parallel to the chord of the parabola joining the points (- 3, 0) & (- 4, 1).
a.
(- 4, 1)
b.
(- 5 2, 1 4)
c.
(- 2, 1)
d.
(- 7 2, 1 4)


Solution:

Slope of the chord joining the points (- 3, 0) & (- 4, 1) = (1 -  0)( - 4 + 3) = - 1

Slope of the tangent to the curve at any point (x, f (x)) is f ′ (x ) = 2(x + 3)

f ′(c) = -1
[By mean value theorem.]

2(c + 3) = -1

c = - 7 / 2 (-4, - 3)

f (c) = (c + 3)2 = (- 72 + 3)2 = 14.

Hence, the point where the tangent to the parabola is parallel to the given chord is (c, f(c)) = (- 7 / 2, 1 / 4).


Correct answer : (4)
 6.  
Find the point on the graph of f (x) = x3, at which the tangent is parallel to the chord joining the points (1, 1) & (3, 27).
a.
((133)12, (133)32)
b.
(13 3, 3 13)
c.
(1, 1)
d.
(0, 0)


Solution:

Slope of the chord joining the points (1, 1) & (3, 27) is = 27 - 13 - 1 = 13

Slope of the tangent to the curve at any point (x, f(x)) is f ′(x) = 3x2.

f ′(c) = 13
[By mean value theorem.]

3 c2 = 13

c = (133)12 (1, 3).

f (c) = (133)32

Hence, the point where the tangent to the curve is parallel to the given chord is (c, f(c)) = ((133)12, (133)32)


Correct answer : (1)
 7.  
Find the point at which the tangent to the curve f(x) = x2 - 6x + 1 is parallel to the chord joining the points (1, - 4) & (3, -8).
a.
(2, -7)
b.
(2, 7)
c.
(-2, -7)
d.
(-2, 7)


Solution:

Slope of the chord joining the points (1, - 4) & (3, - 8) is - 8 + 43 - 1 = - 2

Slope of the tangent to the curve at any point (x, f(x)) is f ′(x) = 2x - 6

f ′(c) = - 2
[By Mean Value Theorem.]

2c - 6 = - 2

c = 2 (1, 3).

f(c) = (2)2 - 6(2) + 1

f(c) = - 7

Hence, the point where the tangent to the curve is parallel to the given chord is (c, f(c)) = (2, - 7)


Correct answer : (1)
 8.  
Find the point on the curve y = 12(x + 1)(x - 2) in the interval [-1, 2], at which the tangent is parallel to the x - axis.
a.
(- 1 2, - 27)
b.
(1 2, - 27)
c.
(- 1 2, 27)
d.
(1 2, 27)


Solution:

y = f(x) = 12(x + 1)(x - 2)
[Given curve.]

= 12x2 - 12x - 24

Slope of the tangent to the curve at any point (x, f (x)) is f ′(x) = 24x -12

f ′ (c) = f(2) - f(-1)2 + 1
[By Mean Value Theorem.]

24c - 12 = 0

c = 1 / 2 (-1, 2)

f(c) = 12(1 / 2)2 - 12(1 / 2) - 24

= 3 - 6 - 24

= - 27

Hence, the point where the tangent to the curve is parallel to x- axis is (c, f (c)) = (1 / 2, - 27).


Correct answer : (2)
 9.  
State whether the function f(x) = ln x on [1, 2] satisfies the mean value theorem.
a.
yes
b.
no


Solution:

f(x) = ln x, x [1, 2]
[Given function.]

f(x) is continuous on [1, 2] and differentiable on (1, 2).

So, there exists c (1, 2) such that f ′ (c) = f(2) - f(1)2 - 1.
[By mean value Theorem.]

1c = ln 2 c = 1ln 2 (1, 2).
[f ′(x) = 1x]

Hence, the given function satisfies the mean value theorem.


Correct answer : (1)
 10.  
State whether the function f (x) = sin x - sin 2x, x [0, π] satisfies the mean value theorem.
a.
yes
b.
no


Solution:

f (x) = sin x - sin 2x, x [0, π]
[Given function.]

The given function is continuous on [0, π] and differentiable in (0, π).

So, there exists c (0, π) such that f ′ (c) = f(π) - f(0)π  - 0.
[By mean value theorem.]

cos c - 2 cos 2 c = 0
[f ′ (x) = cos x - 2 cos 2x.]

cos c = 4 cos2c - 2
[cos 2c = 2 cos2c - 1.]

4 cos2 c - cos c - 2 = 0

c = 32o.53′ , 126o.37′ (0, π) )
[Solve use the calculator.]

Hence, the function f (x) satisfies the mean value theorem.


Correct answer : (1)

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