The value of - in the mean value theorem of fb-fa-b-af- - for fx-Ax2-Bx-C in a-b is

F(x)=Ax^2+Bx+C f'(x)=2Ax+B f'(ξ)=2Aξ +B By mean value theorem, we have f'(ξ)==f(b) f(a)/b a 2Aξ +B=(Ab^2+Bb+C) (Aa^2+Ba+C)/b a 2Aξ+B=A(b+a)+B ξ=b+a/2 nbsp; ...

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Standard X

Mathematics

Arithmetic Mean

The value of ...

Question

The value of $ξ$ in the mean value theorem of $f (b) - f (a) = (b - a) f^{'} (ξ)$ for $f (x) = A x^{2} + B x + C$ in $(a, b)$ is

b + a

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b - a

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\frac{(b + a)}{2}

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\frac{(b - a)}{2}

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Solution

The correct option is C $\frac{(b + a)}{2}$
$f (x) = A x^{2} + B x + C$
$f^{'} (x) = 2 A x + B$
$f^{'} (ξ) = 2 A ξ + B$
By mean value theorem, we have
$f^{'} (ξ) == \frac{f (b) - f (a)}{b - a}$
$2 A ξ + B = \frac{(A b^{2} + B b + C) - (A a^{2} + B a + C)}{b - a}$
$2 A ξ + B = A (b + a) + B$
$ξ = \frac{b + a}{2}$

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The value of ξ in the mean value theorem of f(b)−f(a)=(b−a)f′(ξ) for f(x)=Ax2+Bx+C in (a,b) is

The correct option is C (b+a)2f(x)=Ax2+Bx+C f′(x)=2Ax+B f′(ξ)=2Aξ+B By mean value theorem, we have f′(ξ)==f(b)−f(a)b−a 2Aξ+B=(Ab2+Bb+C)−(Aa2+Ba+C)b−a 2Aξ+B=A(b+a)+B ξ=b+a2

The value of $ξ$ in the mean value theorem of $f (b) - f (a) = (b - a) f^{'} (ξ)$ for $f (x) = A x^{2} + B x + C$ in $(a, b)$ is