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