MHB Solve Multiline Function w/2 Variables - Explanation Here

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To solve a multiline function with two variables, understanding the constraints on the variables is crucial. For instance, if considering f(1 - h) with the condition 0 < h < 1, it follows that 0 < 1 - h < 1, leading to the conclusion that 0 < x < 1. The discussion also touches on the specific function h^2 - 2h - 3, prompting further inquiry about additional functions. Clarification is sought on other related functions beyond the initial example. The conversation emphasizes the importance of defining the function and its parameters for accurate solutions.
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Can anyone explain how to solve a multiline function with two variables? Please see attached.
 

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tobiajj1 said:
Can anyone explain how to solve a multiline function with two variables? Please see attached.
If you are referring to f(1 - h), for example, You know that 0 < h < 1, so you know that 0 < 1 - h < 1 and thus 0 < x < 1. What is f(x)?

-Dan
 
h^2-2h-3?
topsquark said:
If you are referring to f(1 - h), for example, You know that 0 < h < 1, so you know that 0 < 1 - h < 1 and thus 0 < x < 1. What is f(x)?

-Dan
 
tobiajj1 said:
h^2-2h-3?
That's it! What about the other one?

-Dan
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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