|x+2| is continuous everywhere.
for case a,|x+2| is differentiable on (-2,4),whereas for case b,|x+2| is not differentiable on (-3,4) at x=-2..
also the slope of the tangent at every point to the given curve between (-2,0) is 1,which is the same as that of the chord joining the points (-2,0) and (4,6)..
thus all values of x between (-2,4) satisfy lagrange's theorem.
check given function f(x)=Ix+2I verifies lagrange's theorem on :-
a) [-2,4]
b) [-3,4]
c) none of these
here I DENOTES MODULUS....
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3 Answers
Arka Halder
·2010-08-26 09:28:39