calculus
posted by hy .
given h(x) =kx+1, X<1
(k is constant) X^2, x > and equal to 1
A) find value of k such that h(x) is continous at x=1
B) determine if h(x) is differentiable at x=1

a)
For h(x) to be continuous,
we need to address the point x=1, since the rest are continuous.
h(x)=x² for x≥1 means that
h(1)=1²=1
We therefore require that
lim h(x) x>1 = 1, or
set h(x)=kx+1 = 1 => k=0
Therefore h(x)=0*x+1 =1
b) h'(1) = d(1)/dx =0
h'(1)=h'(1+)=d(x²)/dx = 2x = 2
Since h'(1)≠h'(1+), we conclude that h(x) is not differentiable at x=1.