MHB Finding a value that will make a function continuous

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For a function to be continuous at x=2, the values of the function must be equal from both sides at that point. The equation 5(2)-1 = a(2)^2+1 is used to find the value of 'a' that ensures this continuity. Solving this equation yields a = 2, which is the necessary value for continuity. The intuition behind this is that continuity requires the left-hand limit and right-hand limit to match the function's value at that point. Thus, determining 'a' ensures the function behaves consistently at x=2.
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Hi All, just a question regarding continuous functions.
From what I understand if x > 2, then any value of 'a' should make this function continuous? Any clarification would be very helpful!
Thanks in advance!
 

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Try 5(2)-1 = a(2)^2+1 solve for a
 
Thank you, I get a = 2. Can you please explain to me the intuition behind that? Why is it that I have to use 5(2)-1 = a(2)^2+1 and solve for a?
 
For the function to be continuous at x=2, then they must be equal at that point.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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