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.
To find the value of k such that h(x) is continuous at x=1, we need to make sure that the two pieces of the function, kx+1 for x<1 and x^2 for x≥1, meet at x=1 smoothly without any jumps or gaps.
For continuity, we need h(1) to be equal to the limit of h(x) as x approaches 1 from both the left and the right sides.
A) To find k, let's evaluate h(1) and the limits separately for x<1 and x≥1:
For x<1:
h(1) = k(1) + 1 = k + 1
For x≥1:
h(1) = 1^2 = 1
Now, for h(x) to be continuous, h(1) should be equal to the limits of h(x) as x approaches 1 from both sides:
Limit as x approaches 1 from the left (x<1):
lim(x→1-) (kx+1) = k(1) + 1 = k + 1
Limit as x approaches 1 from the right (x≥1):
lim(x→1+) (x^2) = 1^2 = 1
Since both limits are equal to h(1), we can set them equal to each other:
k + 1 = 1
Subtracting 1 from both sides, we get:
k = 0
So, if k equals 0, h(x) will be continuous at x=1.
B) To determine if h(x) is differentiable at x=1, we need to check if the left-hand derivative and the right-hand derivative at x=1 are equal.
The left-hand derivative, h'(1-), is the derivative of kx+1 with respect to x evaluated at x=1. Since k is a constant, the derivative of kx+1 with respect to x is simply k. So, h'(1-) = k.
The right-hand derivative, h'(1+), is the derivative of x^2 with respect to x evaluated at x=1. Taking the derivative of x^2, we get 2x, and evaluating it at x=1, we have h'(1+) = 2(1) = 2.
Since h'(1-) = k and h'(1+) = 2, h(x) is differentiable at x=1 if k = 2.
Therefore, h(x) is differentiable at x=1 when k = 2.