MEAN VALUE THEOREM PRACTICE PROBLEMS AND SOLUTIONS

Using mean value theorem find the values of c.

(1)  f(x)  =  1-x²  [0, 3]

(2)  f(x)  =  1/x, [1, 2]

(3)  f(x)  =  2x³+x²-x-1, [0, 2]

(4)  f(x)  =  x^2/3, [-2, 2]

(5)  f(x)  =  x³-5x² - 3 x       [1 , 3]

(6)  If f(1)  =  10 and f'(x)  ≥ 2 for 1 ≤ x ≤ 4 how small can

f (4) possibly be ?

(7)  At 2.00 p.m car's speedometer reads 30 miles/hr., at 2. 10 pm it reads 50 miles/hr. Show that sometime between 2.00 and 2.10 the acceleration is exactly 120 miles/hr²

Problem 1 :

f(x)  =  1-x²,  [0, 3]

Solution :

1) f(x) is defined and continuous on the interval [0, 3] 

2) f(x) is differentiable on the interval (0,3).

Then there exists at least one point c∊(0, 3) such that 

f'(c)  =  [f(b)-f(a)] / (b-a)

f(x)  =  1-x²

f'(x)  =  -2x

f'(x) = -2x

f'(c)  =  -2c

f(0)  =  1-0² ==>  1

f(3)  =  1-3²  ==> -8

f'(c)  =  [f(b)-f(a)] / (b - a)

2c  =  (-8-1)/(3-0)

2c  =  -3 

c  =  -3/2 ∊ (0, 3)

Problem 2 :

f(x)  =  1/x, [1, 2]

Solution :

If f (x) be a real valued function that satisfies the following conditions.

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

2. f(x) is differentiable on (1, 2).

Then there exists at least one point c ∊ (1,2) such that 

f'(c)  =  [f(b)-f(a)] / (b-a)

f(x)  =  1/x

f(x)  =  x^-1

f'(x)  =  (-1) x^-2

f'(x)  =  - 1/x²

f'(c)  =  - 1/c²

f(1)  =  1/1  ==>  1

f(2)  =  1/2

f'(c)  =  [f(b)-f(a)] /(b-a)

=  (-1/2)/1

=  (-1/2)

-1/c²  =  -1/2

c²  =  2

c  =  ±√2

c  =  √2, -√2

-√2 ∉ (1, 2)  but  √2 ∈ (1, 2).

So the required value of c is √2.

Problem 3 :

f(x)  =  2x³+x²-x-1, [0, 2]

Solution :

If f (x) be a real valued function that satisfies the following conditions.

1) f(x) is defined and continuous on [0, 2] 

2) f(x) is differentiable on (0, 2).

Then there exists at least one point c ∊ (0, 2) such that 

f'(c)  =  [f(b)-f(a)]/(b-a)

f(x)  =  2x³+x²-x-1

f'(x)  =  6x²+2x-1

f'(c)  =  6c²+2c-1

f(0)  =  6(0)²+2(0)-1

f(0)  =  -1

f(2)  =  2x³ + x² - x - 1

=  2(2)³+2²-2-1

=  2(8)+4-3

=  16+4-3

=  17

f'(c)  =  [f(b)-f(a)] / (b-a)

=  [17-(-1)]/(2 - 0)

=  [17+1]/2

=  9

6c²+2c-1  =  9 

6c²+2c-1-9  =  0

6c²+2c-10  =  0

3c²+c-5  =  0

We cannot solve this equation by using factorization method. So we are trying to solve this equation by using quadratic formula.

a  =  3, b  =  1, c  =  -5     

x  =  -1±√[1² - 4(3)(-5)]/2(3)

x = -2± √[1 + 60]/6

x  =  -2±√61/6

x  =  (-2+√61)/6,  (-2-√61)/6

c  =  (-2+√61)/6 ∈ (0, 2)  L

Problem 4 :

f(x)  =  x^2/3, [-2, 2]

Solution :

If f (x) be a real valued function that satisfies the following conditions.

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

2) f(x) is not differentiable on (-2, 2).

So, mean value theorem does not exists.

Problem 5 :

f(x)  =  x³-5x² - 3 x       [1 , 3]

Solution :

If f(x) be a real valued function that satisfies the following conditions.

1) f(x) is defined and continuous on [1, 3] 

2) f(x) is differentiable on (1,3).

Then there exists at least one point c ∊ (1,3) such that 

f'(c)  =  [f(b)-f(a)]/(b - a)

f(x)  =  x³-5x²-3x

f'(x)  =  3x²-5(2x)-3

f'(x)  =  3x²-10x-3

f'(c)  =  3c²-10c-3  ---(1)

f(1)  =  x³-5x²-3x

=  1³-5(1)²-3(1)

=  1-5-3

=  -7

f(1)  =  -7

f(3)  =  x³-5x²-3x

=  3³-5(3)²-3(3)

=  27-45-9

= 27- 54

=  -27

f(3)  =  -27

f'(c) = [f(b)-f(a)] / (b-a)

=  [-27-(-7)]/(3-1)

=  [-27+7]/2

=  -20/2

[f(b)-f(a)] / (b-a)  = -10 ---(2)

(1)  =  (2)

3c²-10c-3  = -10 

3c²-10c-3+10  =  0

3c²-10c+7  =  0

(c-1) (3c-7)  =  0

c  =  1 and c  =  7/3.

Problem 6 :

If f(1)  =  10 and f'(x)  ≥ 2 for 1 ≤ x ≤ 4 how small can f (4) possibly be ?

Solution :

If f (x) be a real valued function that satisfies the following conditions.

1)  f(x) is defined and continuous on [1, 4] 

2)  f(x) is differentiable on (1, 4).

Then there exists at least one point c ∊ (1, 4) such that

f'(c)  =  [f(b)-f(a)] / (b-a)

f'(c)  =  [f(4)-f(1)] / (b-a)

f'(c)  =  [f(4)-10]/(4-1)

f'(c)  =  [f(4)-10]/3

3f'(c)  =  [f (4)-10]

3f'(c)+10  =  f(4)

f(4)  ≥  6+10

f(4) ≥ 16.

So the minimum value of f(4) must be 16.

Problem 7 :

At 2.00 p.m car's speedometer reads 30 miles/hr., at 2. 10 pm it reads 50 miles/hr. Show that sometime between 2.00 and 2.10 the acceleration is exactly 120 miles/hr²

Solution :

Let velocity be  v at time t.

1) f(x) is defined and continuous on [2, 2.10] 

2) f(x) is differentiable on (2, 2.10).

v (2)  =  30 miles/hour

v(2.10)  =  50 miles/hour

we can find the value of c by using the above condition

v'(c)  =  [v(2.10)-v(2)]/2.10 - 2

=  [50 - 30]/(10/60) miles/hour

=  20/(10/60) miles/hour

=  20 x (60/10)

=  120 miles/hour

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.