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what is the value of constant C in the Langrange mean value theorem,if f(x)=2x^2+3x+4 in [1,2]?

what is the value of constant C in the Langrange mean value theorem,if f(x)=2x^2+3x+4 in [1,2]?

Grade:Upto college level

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
f'(c) = \frac{f(b)-f(a)}{b-a}
f'(c) = \frac{f(2)-f(1)}{2-1}
f'(c) = \frac{18-9}{2-1} = 9
f(x) = 2x^{2}+3x+4
f'(x) = 4x+3 = 9
\Rightarrow x = \frac{3}{2}
\Rightarrow c = \frac{3}{2}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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