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f(X)=(1-x)*(2-x) for x lies between 1 and 2 inclusive;f(x)=(3-x) for x>2 ; Find whether x is differentiable at x=2 ; By { f(x+h)-f(x) }/h as h tends to zero method ,left hand derivate is 1 and right hand derivative is infinite; so at x=2 it is not differentiable by differentiation method ,i got right hand derivative=-1 , but using above method i got infinite.but Right hand limit by both the methods must be same ..Can anyone tell me why is this so ?
we see that lim x tends to 2- f(x)= lim x tends to 2- (1-x)*(2-x) = 0while lim x tends to 2+ f(x)= lim x tends to 2+ (3-x) = 1obviously RHL does not equal LHL. so it is not continuous.recall the theorem that if f(x) is not continuous at x=a, it cannot be differentiable too at x=a.so f(x) is not differentiable at 2 since it is not continuous at x= 2 in the first place.kindly approve yar :=)
Der student lets use the { f(x+h)-f(x) }/h method for right hand limit. we wish to find lim h tends to 0+ { f(2+h)-f(2) }/hnow, you have typed the question incompletely. so i dont know what f(2) is. actually, its not ur fault it is askiitians websites fault that when u try to type inequality signs, they often mess up.so, kindly reupload the ques with a pic.btw, it doesnt matter what the actual ques is. the fact is that f(x) is not differentiable at x=2. the reason is that f(x) is not continuous at x=2.we see that lim x tends to 2- f(x)= lim x tends to 2- (1-x)*(2-x) = 0while lim x tends to 2+ f(x)= lim x tends to 2+ (3-x) = 1so, no matter what the value of f(2) is, f(x) can never be differentiable at 2 since it is not continuous there in the first place.
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