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evablogger [386]
3 years ago
8

Can anybody please help me out how solved this problem pleased

Mathematics
1 answer:
maks197457 [2]3 years ago
3 0

Answer:

i am sevn i ned hlp wht 1 plas 3 plz hlp

Step-by-step explanation:

I ned hlp

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For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that a is such that

\displaystyle\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)\iff\lim_{x\to0}ax=\lim_{x\to0}x^2-3x

Obviously, both limits are 0, so g is indeed continuous at x=0.

Now, for g to be differentiable everywhere, its derivative g' must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals:

g'(x)=\begin{cases}a&\text{for }x0\end{cases}

and at this time, we don't know what's going on at x=0, so we omit that case. We want g' to be continuous, so we require that

\displaystyle\lim_{x\to0^-}g'(x)=\lim_{x\to0^+}g'(x)\iff\lim_{x\to0}a=\lim_{x\to0}2x-3

from which it follows that a=-3.

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Steve drives 5/8 miles on 1/2 gallon of gas. Find the miles per gallon.
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