Posted by **hy** on Monday, October 25, 2010 at 12:03am.

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

- calculus -
**MathMate**, Monday, October 25, 2010 at 8:47am
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.

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