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tangare [24]
3 years ago
13

I don’t know how to do this

Mathematics
1 answer:
stich3 [128]3 years ago
4 0

To check for continuity at the edges of each piece, you need to consider the limit as x approaches the edges. For example,

g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}

has two pieces, 2x+5 and x^2-10, both of which are continuous by themselves on the provided intervals. In order for g to be continuous everywhere, we need to have

\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)

By definition of g, we have g(-3)=2(-3)+5=-1, and the limits are

\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1

\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1

The limits match, so g is continuous.

For the others: Each of the individual pieces of f,h are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.

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Step-by-step explanation:

Consider the provided information.

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ik the doesnt rlly make any sense but hope this helps

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