Q. Leth(x)={x+56x2−6 for −5<x<−3 for x≥−3Is h continuous at x=−3 ?Choose 1 answer:(A) Yes(B) No
Check Left-Hand Limit: To determine if h(x) is continuous at x=−3, we need to check if the left-hand limit as x approaches −3, the right-hand limit as x approaches −3, and the value of the function at x=−3 are all equal.
Calculate Left-Hand Limit: First, we calculate the left-hand limit of h(x) as x approaches −3 from the left. This means we use the first piece of the function, which is h(x)=(x+5)6 for -5 < x < -3.
Calculate Right-Hand Limit: The left-hand limit is the limit of (x+5)6 as x approaches −3 from the left. We substitute x with values that are slightly greater than −3 and see if the limit approaches a specific value.limx→−3−(x+5)6=(−3+5)6=26=3
Check Right-Hand Limit: Next, we calculate the right-hand limit of h(x) as x approaches −3 from the right. This means we use the second piece of the function, which is h(x)=x2−6 for x≥−3.
Check Value at x=−3: The right-hand limit is the limit of x2−6 as x approaches −3 from the right. We substitute x with −3 directly since the function is a polynomial and thus continuous.limx→−3+x2−6=(−3)2−6=9−6=3
Confirm Continuity: Now, we need to check the value of the function at x=−3. Since x=−3 falls in the interval for the second piece of the function, we use h(x)=x2−6.h(−3)=(−3)2−6=9−6=3
Confirm Continuity: Now, we need to check the value of the function at x=−3. Since x=−3 falls in the interval for the second piece of the function, we use h(x)=x2−6.h(−3)=(−3)2−6=9−6=3Since the left-hand limit, the right-hand limit, and the value of the function at x=−3 are all equal to 3, the function h(x) is continuous at x=−3.