IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

cauchy089

Let f(x) = (x+1).|x^2-1| , -(infinity)<x< +(infinity)

verify the differentiability of the function f(.) at the point x=-1 and x=1

Let f(x)=3(x-2)^2/3 - (x-2) , x belongs to [0,20]

Let x' and y' be the point of global minima and the global maxima,respectively,of f(.) in the

interval [0,20]

evalute f( x' ) + f( y' )

iitkgp_bipin

f(x) = (x+1).|x²-1| = (x+1).|(x+1)(x-1)|

When x< -1 and x>1 , (x+1)(x-1)>0
|(x+1)(x-1)| = (x+1)(x-1)

When -1<=x<=1 , (x+1)(x-1) <= 0
|(x+1)(x-1)| = -(x+1)(x-1)

f(x) = (x+1)²(x-1) for x< -1
= -(x+1)²(x-1) for -1<=x<=1
= (x+1)²(x-1) for x>1

At x = -1 :
LHD = _[h]

_[0] {f(-1-h) - f(-1)}/(-h) = _[h]

_[0] {(-h)²(-2-h)}/(-h) = 0
RHD = _[h]

_[0] {f(-1+h) - f(-1)}/h = _[h]

_[0] {-(h)²(-2+h)}/h = 0
Since LHD = RHD at , f(x) is differentiable at x = -1.

At x = 1 :
LHD = _[h]

_[0] {f(1-h) - f(1)}/(-h) = _[h]

_[0] {-(2-h)²(-h)}/(-h) = -4
RHD = _[h]

_[0] {f(1+h) - f(1)}/h = _[h]

_[0] {(2+h)²(h)}/h = 4
Since LHD

RHD , f(x) is not differentiable at x = 1.

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