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+1 vote
Is following function Satisfy Lagrange's Mean Value Theroem?
f(x) = | x+2 |  in [-2, 0]

Detailed solution PLEASE!
asked in Calculus by Veteran (11.4k points)   | 106 views
Change your comment to answer @Akriti sood it correct?

2 Answers

+4 votes
Best answer

for lagrange's theorem,

1) it should be continous at [-2,0]

f(x) = |x+2| =-(x+2 ) x< -2

(x+2)   -2<= x< -1

(x+2)   -1< x < 0

(x+2)    x>0


so,for continouty check limit at -2  lim x ->-2- -(x+2) =0

rhs at -2------ lim x-> -2(x+2) =0

and f(-2) =0

hence,continous at x=-2

similarly,it is continous at -1 and 0

2) f(x) should be differentiable at x= ]-2,0[

for checking diferentibilty

lim x-> c- f(x) -f(c)/x-c =lim x->cf(x) -f(c)/x-c

which is same as f(x) for lhs and rhs of -1 is same.

it should  follow lagranje



answered by Veteran (10.1k points)  
selected by
yes...but only satisfying continuity and differentiabilty doesnt guarantee u that u could always find such c in (a,b) right?
u should check for c also before saying whether lagranges will be applicable or nt!!
Akriti is correct here, we just need to check first 2 conditions to check if lagrange's theroem can be applied or not
@vijay means every continuous and differentiable fn for a range is applicable for lagrange's mean value theorem?
yes,if given function is continuous and differentiable, there will be such C definitely.
lagranges theorem means that, if a curve is smooth beween [a,b] and no sharp edges  then there will be a point "c" in (a,b) such that slope of line joining f(a),f(b) is equal to slope of tangent drawn at point C.
+1 vote
you can observe by substituting -2,-1,0 in the given expression that its a straight line joining (-2,0) and (0,2) in the 2nd quadrant.
its a straightline so its differentiable and continous.
so lLagrange's MVT is applicable
answered by Boss (9.4k points)  
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