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lawyer [7]
3 years ago
5

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
5 0
Assuming the function is

f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}

For f(x) to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)
\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)
\iff2=b+c

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

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

\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)
\iff2=2b+c

Now we solve for b and c:

\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2
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