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Find the values of $a$ and $b$ if the following function is differentiable at $x=1$.

$$ f(x)=\left\{ \begin{aligned} x^2 + 3x + a & & x\le1, \\ bx + 2~~~~~~~~~ & & x>1. \end{aligned} \right. $$

Thanks in advance for the help!

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    What did you try? – 2017-01-13
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    I tried equating the left-hand derivatives and right-hand derivatives. Moreover, i had some confusion. – 2017-01-13

2 Answers 2

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For a function to be differentiable at a point it's value at that point has to equal limit of that function at that point and derivatives from both sides must be the same. So in this case:\begin{aligned}x^2 + 3x + a = bx + 2\end{aligned}\begin{aligned}2x + 3 = b\end{aligned} By plugging in 1 you can easily get that a equals 3, and b equals 5.

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    Thanks, I was a bit confused before. – 2017-01-13
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It is enough to make one-sided limits equal to each other (for continuity purposes) and make one-sided derivatives equal to each other. So we have the conditions: $4+a=b+2$ (for limits) and $b=5$ (for derivatives). Then $a=3,b=5$.