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lesya [120]
3 years ago
14

Determine the values of the constants B and C so that the function given below is differentiable.

Mathematics
1 answer:
laila [671]3 years ago
5 0
For the function to be differentiable, its derivative has to exist everywhere, which means the derivative itself must be continuous. Differentiating gives

f'(x)=\begin{cases}24x^2&\text{for }x1\end{cases}

The question mark is a placeholder, and if the derivative is to be continuous, then the question mark will have the same value as the limit as x\to1 from either side.

\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}24x^2=24
\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}B=B

So the derivative will be continuous as long as B=24

For the function to be differentiable everywhere, we need to require that f(x) is itself continuous, which means the following limits should be the same:

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}8x^3=8
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}Bx+C=24+C

24+C=8\implies C=-16

So, the function should be

f(x)=\begin{cases}8x^3&\text{for }x\le1\\24x-16&\text{for }x>1\end{cases}

with derivative

f'(x)=\begin{cases}24x^2&\text{for }x
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