ANSWER: k=5/2
EXPLANATION. On the set [1,2) the function F(x)=3-kx is continuous since it is elementary.
On the set "\\quad \\left( 2,+\\infty \\right) \\quad F(x)=\\frac { { x }^{ 2 } }{ 4 } -3" is continuous since it is elementary.
At the point x=2 the function is continuous if
"\\lim _{ x\\rightarrow 2-0 }{ F(x)= } \\lim _{ x\\rightarrow 2+0 }{ F(x)= } F(2)=\\frac { { 2 }^{ 2 } }{ 4 } -3=-2"
"\\lim _{ x\\rightarrow 2-0 }{ F(x)= } \\lim _{ x\\rightarrow 2-0 }{ \\left( 3-kx \\right) } =3-2k,\\\\ \\lim _{ x\\rightarrow 2+0 }{ F(x)= } \\lim _{ x\\rightarrow 2+0 }{ \\left( \\frac { { x }^{ 2 } }{ 4 } -3 \\right) = } 1-3=-2=F(2)\\quad"
Therefore, if 3-2k=-2 i.e. k=5/2 then the function is continuous at the point x=2. For other values of k , the function is discontinuous at this point.
Comments
Leave a comment