Let h be defined by

h(x)=f(x)*g(x) x less than or equal to 1

h(x)=k+x if x > 1

If lim as x approaches 1 f(x)=2 and lim as x approaches 1 g(x)=-2 then for what value of k is h continous?

A. -5
B. -4
C. -2
D. 2

Is it a

yes

To determine the value of k for which h is continuous, we need to make sure that h(x) is continuous at x = 1.

For a function to be continuous at a particular point, three conditions need to be satisfied:
1. The function exists at that point.
2. The limit of the function as x approaches that point exists.
3. The value of the function at that point is equal to the limit.

Let's check the three conditions for h(x) at x = 1:

1. The function exists at x = 1, as there is a specific rule defined for h(x) when x > 1.

2. The limit of f(x) as x approaches 1 is given to be 2, and the limit of g(x) as x approaches 1 is given to be -2. To find the limit of h(x) as x approaches 1, we substitute these values into the rule for h(x) when x ≤ 1:

lim(x→1-) h(x) = lim(x→1-) (f(x) * g(x)) = (lim(x→1-) f(x)) * (lim(x→1-) g(x)) = 2 * (-2) = -4

Therefore, the limit of h(x) as x approaches 1 from the left is -4.

3. Now, let's check if the value of h(x) at x = 1 is equal to the limit from the left:
h(1) = k + 1 (from the rule defined for x > 1)

For h(x) to be continuous, the value of h(x) at x = 1 should be equal to the limit from the left, which is -4:
k + 1 = -4

Solving this equation, we find that k = -5.

Therefore, the value of k for which h(x) is continuous is -5.

Answer: A. -5