# Verify mean value theorem for each of the functions given

Given:

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

Firstly, we have to show that f(x) is continuous.

Here, f(x) is continuous because f(x) has a unique value for each x [1,5]

Condition 2:

Now, we have to show that f(x) is differentiable

[using chain rule]

f’(x) exists for all x (1,5)

So, f(x) is differentiable on (1,5)

Hence, Condition 2 is satisfied.

Thus, mean value theorem is applicable to given function.

Now,

Now, we will find f(a) and f(b)

so, f(a) = f(1)

and f(b) = f(5)

Now, let us show that c (1,5) such that

On differentiating above with respect to x, we get

Put x = c in above equation, we get

By Mean Value theorem,

Squaring both sides, we get

16c2 = 24 × (25 – c2)

16c2 = 600 – 24c2

24c2 + 16c2 = 600

40c2 = 600

c2 = 15

c = √15 (1,5)

Hence, Mean Value Theorem is verified.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Super 10 Question: Check Your Knowledge of Maxima & Minima (Quiz)45 mins
Maxima & Minima in an interval60 mins
Questions Based on Maxima & Minima57 mins
Connection B/w Continuity & Differentiability59 mins
Questions based on Maxima & Minima in an interval59 mins
Check your Knowlege of Maxima & Minima ( Challenging Quiz)60 mins
Problems Based on L-Hospital Rule (Quiz)0 mins
When does a Maxima or Minima occur?48 mins
Interactive Quiz | Differentiability by using first principle59 mins
Interactive Quiz on Limits67 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses