Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
532 views
in Continuity and Differentiability by (51.9k points)

Verify Lagrange’s mean value theorem for the functions on the indicated intervals. Find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem:

f(x) = x2 + x – 1 on [0, 4]

1 Answer

+1 vote
by (50.8k points)
selected by
 
Best answer

Given as f(x) = x2 + x – 1 on [0, 4]

The every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Since, f(x) is a polynomial function. Therefore it is continuous in [0, 4] and differentiable in (0, 4). Therefore both the necessary conditions of Lagrange’s mean value theorem is satisfied. So, there exist a point c ∈ (0, 4) such that:

f'(c) = (f(4) - f(0))/(4 - 0)

f'(c) = (f(4) - f(0))/4

f (x) = x2 + x – 1

Differentiate with respect to x:

f’(x) = 2x + 1

For the f’(c), put the value of x = c in f’(x):

f’(c) = 2c + 1

For the f(4), put the value of x = 4 in f(x):

f (4) = (4)2 + 4 – 1

= 16 + 4 – 1

= 19

For the f(0), put the value of x = 0 in f(x):

f (0) = (0)2 + 0 – 1

= 0 + 0 – 1

= – 1

Thus, lagrange's mean value theorem is verified.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...