Question

For what value of is the function defined by continuous at x = 0? What about continuity at x = 1?

Answer 1

The given function is,

f( x )={ λ( x 2 −2x )   ,  x≤0 4x+1           ,  x>0

The left hand limit of the function is,

LHL= lim x→ 0 − f( x ) = lim x→ 0 − λ( x 2 −2x ) =λ[ 0 2 −2( 0 ) ] =0

The right hand limit of the function is,

RHL= lim x→ 0 + f( x ) = lim x→ 0 + ( 4x+1 ) =4(0)+1 =1

As the function is continuous at x=0, so LHL=RHL.

lim x→ 0 − f( x )= lim x→ 0 + f( x ) 0=1

This is not possible.

Therefore, there is no real value of λ for which the given function is continuous at x=0.

When x=1, the function becomes,

f( 1 )=4( 1 )+1 =5

The limit of the function is,

lim x→1 f( x )= lim x→1 4( 1 )+1 =5

It can be observed that, lim x→1 f( x )=f( 1 ).

Therefore, the function is continuous for all real values of λ at x=1.