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Consider the following problem:

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I got $g'(1) = f'(1) = 2$, which is correct according to the expected answer. However, I get that $a$ can be any value, as it doesn't affect anything and $g$ is differentiable no matter what $a$ is (am I missing something here?). The expected answer is that $a$ must be $2$ however.

Can someone explain why $a$ needs to be $2$?

2 Answers 2

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The derivative of g from the right at $x=1$ is a, and from the left it's $f^\prime(1) = 2$. They must be equal.

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    Right, I thought it was $f(x)$ not $f(1)$ at one point. Makes sense, thanks.2010-09-24
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    By the way, that $f(1)$ is of course there in order to make $g$ continuous at $x=1$. (For the argument to be complete, this needs to be checked as well, since if $g$ is not continuous, it has no chance of being differentiable. In particular, it is not enough to merely argue that $(d/dx)(ax-a+f(1))=a$, since if the right-hand limit of $g$ at $x=1$ is different from $g(1)$, then the derivative from the right will be $\infty$ or $-\infty$, not $a$!)2010-09-24
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Hint: if $g$ is supposed to be differentiable at $x=1$, the left and right derivatives of $g$ should be identical. Now, find the value of $a$ that would make it so.