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If the following function is continuous, then what is the value of b?๐‘”(๐‘ก)={2๐‘ก2+2๐‘กโˆ’24๐‘กโˆ’3๐‘–๐‘“ย ๐‘กโ‰ 3๐‘๐‘–๐‘“ย ๐‘ก=3g(t)={ tโˆ’32t 2 +2tโˆ’24โ€‹ bโ€‹ ifย tโ‰ 3ifย t=3โ€‹ A.14B.3C.7D.None of theseE.0

Question

If the following function is continuous, then what is the value of b?๐‘”(๐‘ก)={2๐‘ก2+2๐‘กโˆ’24๐‘กโˆ’3๐‘–๐‘“ย ๐‘กโ‰ 3๐‘๐‘–๐‘“ย ๐‘ก=3g(t)={ tโˆ’32t 2 +2tโˆ’24โ€‹ bโ€‹ ifย tโ‰ 3ifย t=3โ€‹ A.14B.3C.7D.None of theseE.0

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Solution

The function g(t) is said to be continuous at t=3 if the limit of g(t) as t approaches 3 is equal to g(3).

Given the function g(t) = 2t^2 + 2t - 24 / t - 3 for t โ‰  3 and g(t) = b for t = 3, we first need to find the limit of g(t) as t approaches 3.

The expression 2t^2 + 2t - 24 can be factored to 2(t - 3)(t + 4). So, g(t) = 2(t - 3)(t + 4) / (t - 3) for t โ‰  3.

For t โ‰  3, the (t - 3) terms cancel out, leaving g(t) = 2(t + 4) = 2t + 8.

Now, we find the limit as t approaches 3: lim (t->3) (2t + 8) = 2*3 + 8 = 14.

Since the function is continuous at t=3, g(3) should also be 14. Therefore, b = 14.

So, the answer is A.14.

This problem has been solved

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