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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 (22.9k points) 48 218 361
edited by | 394 views

1 Answer

+3 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 (3.5k points) 2 12 37
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