If the following function is continuous, then what is the value of b?๐(๐ก)={2๐ก2+2๐กโ24๐กโ3๐๐ย ๐กโ 3๐๐๐ย ๐ก=3g(t)={ tโ32t 2 +2tโ24โ 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(๐ก)={2๐ก2+2๐กโ24๐กโ3๐๐ย ๐กโ 3๐๐๐ย ๐ก=3g(t)={ tโ32t 2 +2tโ24โ bโ ifย tโ 3ifย t=3โ A.14B.3C.7D.None of theseE.0
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.
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