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White raven [17]
3 years ago
12

Find the values of m and b that make the following function differentiable.

Mathematics
1 answer:
KengaRu [80]3 years ago
7 0

f(x)=\begin{cases}x^2&\text{for }x\le2\\mx+b&\text{for }x>2\end{cases}

In order to be differentiable everywhere, f must first be continuous everywhere, which means the limits from either side as x\to2 must be the same and equal to f(2). By definition, f(2)=2^2=4, and

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}x^2=4

\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(mx+b)=2m+b

so we need to have 4m+b=4.

For f to be differentiable at x=2, the derivative needs to be continuous at x=2, i.e.

\displaystyle\lim_{x\to2^-}f'(x)=\lim_{x\to2^+}f'(x)

We then need to have

\displaystyle\lim_{x\to2}2x=\lim_{x\to2}m\implies\boxed{m=4}

Then

2m+b=4\implies\boxed{b=-4}

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Hey there :)

We know the distance formula is:
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1) Coordinates are: ( -10, -5 ) , ( 8 , 5 )
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We do the same for the rest

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