MHB Troubleshooting Inequality: Can't Seem to Solve the Last One

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The discussion focuses on solving an inequality involving a function f(x) and determining the correct intervals where f(x) is less than zero. The user is struggling to identify the last interval needed for the solution. It is clarified that the values -2 and 3 cannot be included in the solution since f(x) must be strictly less than zero. The correct intervals are identified as (-∞, -2) and (3, ∞). The conversation emphasizes the importance of understanding strict inequalities in interval notation.
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Trying to get the last one, but not able to do it. What am I missing. I know that I can't but [-2,infinity] or [3,infinity] because for -2, and 3 y=0 and is not > 0. Any help appreciated.
 

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Is g(-2) > 0? No, so that part of the interval will be [math]( -\infty, -2)[/math].

-Dan
 
Also, because the question requires that f(x) be strictly less than 0, f(x)< 0 you cannot include x= -2 or x= 3. f(x)< 0 is true for (-\infty, -2)\cup (3, \infty).
 
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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