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Let

h(x)={[(6)/(x+5)," for "-5 < x < -3],[x^(2)-6," for "x >= -3]:}
Is 
h continuous at 
x=-3 ?
Choose 1 answer:
(A) Yes
(B) No

Let\newline\( h(x)=\left\{\begin{array}{cl}\frac{6}{x+5} & \text { for }-5

Full solution

Q. Let\newlineh(x)={6x+5 for 5<x<3x26 for x3 h(x)=\left\{\begin{array}{cl}\frac{6}{x+5} & \text { for }-5<x<-3 \\ x^{2}-6 & \text { for } x \geq-3\end{array}\right. \newlineIs h h continuous at x=3 x=-3 ?\newlineChoose 11 answer:\newline(A) Yes\newline(B) No
  1. Check Left-Hand Limit: To determine if h(x)h(x) is continuous at x=3x = -3, we need to check if the left-hand limit as xx approaches 3-3, the right-hand limit as xx approaches 3-3, and the value of the function at x=3x = -3 are all equal.
  2. Calculate Left-Hand Limit: First, we calculate the left-hand limit of h(x)h(x) as xx approaches 3-3 from the left. This means we use the first piece of the function, which is h(x)=6(x+5)h(x) = \frac{6}{(x+5)} for -5 < x < -3.
  3. Calculate Right-Hand Limit: The left-hand limit is the limit of 6(x+5)\frac{6}{(x+5)} as xx approaches 3-3 from the left. We substitute xx with values that are slightly greater than 3-3 and see if the limit approaches a specific value.\newlinelimx36(x+5)=6(3+5)=62=3\lim_{x\to-3^-} \frac{6}{(x+5)} = \frac{6}{(-3+5)} = \frac{6}{2} = 3
  4. Check Right-Hand Limit: Next, we calculate the right-hand limit of h(x)h(x) as xx approaches 3-3 from the right. This means we use the second piece of the function, which is h(x)=x26h(x) = x^2 - 6 for x3x \geq -3.
  5. Check Value at x=3x = -3: The right-hand limit is the limit of x26x^2 - 6 as xx approaches 3-3 from the right. We substitute xx with 3-3 directly since the function is a polynomial and thus continuous.\newlinelimx3+x26=(3)26=96=3\lim_{x\to-3^+} x^2 - 6 = (-3)^2 - 6 = 9 - 6 = 3
  6. Confirm Continuity: Now, we need to check the value of the function at x=3x = -3. Since x=3x = -3 falls in the interval for the second piece of the function, we use h(x)=x26h(x) = x^2 - 6.\newlineh(3)=(3)26=96=3h(-3) = (-3)^2 - 6 = 9 - 6 = 3
  7. Confirm Continuity: Now, we need to check the value of the function at x=3x = -3. Since x=3x = -3 falls in the interval for the second piece of the function, we use h(x)=x26h(x) = x^2 - 6.h(3)=(3)26=96=3h(-3) = (-3)^2 - 6 = 9 - 6 = 3Since the left-hand limit, the right-hand limit, and the value of the function at x=3x = -3 are all equal to 33, the function h(x)h(x) is continuous at x=3x = -3.

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