MHB Finding Constants for a Piecewise Defined Function

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To find the constants a and b for a piecewise defined function, one must ensure continuity at the point of interest, x = 6. The correct approach involves setting the left-hand limit equal to the right-hand limit, specifically using the equations lim_{x→6-} (20) = lim_{x→6+} (8x + a). This ensures that the function does not have any jumps or breaks at x = 6. The discussion emphasizes the importance of continuity in defining piecewise functions. Proper notation for mathematical expressions is also highlighted for clarity.
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hi(Smile)
I need some start help finding the two constants a and b:

Do i start like this?:
$$\lim_{{x}\to{6-}} (20) = \lim_{{x}\to{6+}} (8x+a)$$View attachment 8782
 

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Hi, and welcome to the forum.

Do i start like this?:
\lim_{{x}\to{6-}} (20) = \lim_{{x}\to{6+}} (8x+a)
If you need the function to be continuous, then yes.

You can typeset your formulas by enclosing them in $$...$$ tags (button with $\sum$ on the toolbar) or dollar signs.
 
Evgeny.Makarov said:
Hi, and welcome to the forum.

If you need the function to be continuous, then yes.

You can typeset your formulas by enclosing them in $$...$$ tags (button with $\sum$ on the toolbar) or dollar signs.

Thank you(Smile)(Yes)
 
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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