GATE CSE
First time here? Checkout the FAQ!
x
0 votes
245 views

Question:

The value of $\zeta$ of  $f(b) - f(a) = (b-a) \cdot f'(\zeta)$ for the function $f(x) = Ax^2 + Bx +C$ in the interval $[a,b]$ is _______

 

My Attempt :

STEP 1: $f(x)$ is polynomial function therefore continuous

STEP 2: $f(x)$ is polynomial function therefore differentiable 

STEP 3: There exists a  $c$ such that  $a \leq c \leq b$, such that

$$\begin{align}
f'(c) &= \frac{f(b) - f(a)} {b - a} \\[1em]
f'(c) &= 2Ax + B \\[1em]
\hline
f(b) - f(a) &= (b - a) \cdot f'(\zeta) \\[1em]
2Ax+B &= \frac{(b - a) \cdot f'(\zeta)} {b - a}
\end{align}$$

How to proceed further?

asked in Calculus by Veteran (20.2k points)  
edited by | 245 views

1 Answer

+4 votes
Best answer

Given  f(x) = Ax+ Bx +C, Domain  = [a,b]

This function is  polynomial function,so it is continuous & differentiable in its domain [a,b].

Hence LMVT(Lagrange's Mean Value Theorem) is applicable here.

Let there exists a  'ζ'  such that a < ζ < b, So from LMVT ,we have,

f(b) - f(a) /(b-a)  = f'(ζ)  => [(Ab+Bb +C) - (Aa+Ba +C) ]/(b-a)  = 2Aζ + B

=> [A(b2 - a2) + B(b-a)]/(b-a)  = 2Aζ + B

Since b & a are not equal so cancelling (b-a) from Numerator & denominator,we get,

=>A(b+a) + B  = 2Aζ + B  => ζ = (b+a)/2

Hence,  ζ = (b+a)/2

 

answered by Loyal (3.4k points)  
selected by

Related questions

+3 votes
1 answer
1
asked in Calculus by pC Veteran (20.2k points)   | 137 views
+1 vote
1 answer
2


Top Users Mar 2017
  1. rude

    5236 Points

  2. sh!va

    3054 Points

  3. Rahul Jain25

    2920 Points

  4. Kapil

    2732 Points

  5. Debashish Deka

    2602 Points

  6. 2018

    1574 Points

  7. Vignesh Sekar

    1430 Points

  8. Bikram

    1424 Points

  9. Akriti sood

    1420 Points

  10. Sanjay Sharma

    1128 Points

Monthly Topper: Rs. 500 gift card

21,549 questions
26,889 answers
61,249 comments
23,251 users