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I'm wondering if someone can just take a look at my working and tell me if I've done anything wrong/forgotten to answer some part. The question is as below:

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The limits of the function as X approaches 0 from the left is 5, so in order to make this function continuous we assign a = 2 to make the function as X approaches 0 from the right also = 5.

For differentiability, I've done the below:

for lim (x+2)^2 + 1 (as a = 2) as x appproaches 0


the derivative will be (f(x) - f(0))/x-0


Which is equal to: lim(x->0+) ((x+2)^2 + 1 - 5)/x


as we recall f(0) = 5 as above, we expand:

(x^2 + 4x - 4 + 4) = x + 4 = 4

Therefore the derivative of this function from the right is 4.

for lim (2x+5) as x appproaches 0


the derivative will be (f(x) - f(0))/x-0

Recall f(0) = 5, therefore:

lim(x->0-) ((2x+5)-5)/x = 2

Therefore this function is not differentiable.

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  • $\begingroup$ I guess it's all correct. It ain't differentiable for$a=2$. $\endgroup$ – The Dead Legend Apr 23 '17 at 0:51
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    $\begingroup$ Did you try $a=-2$? $\endgroup$ – Paul Sundheim Apr 23 '17 at 1:06
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  1. $f$ is continuous at $0$ if and only if $$\lim_{x \to 0^+}f(x)=\lim_{x \to 0^-}f(x)\\\iff \lim_{x \to 0^+} x^2+2ax+a^2+1=\lim_{x \to 0^-}2x+5 \\\iff a^2+1=5\iff a^2=4 \iff a =\pm2$$
  2. You reassonning is correct for $a=2$, you can try to apply the same reasonning for $a=-2$.
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