Find $\alpha$ and $\beta$ so that $f(x)$ is continuously differentiable The function $f(x)$ is defined as following 
$$
f(x) := \begin{cases}
\cos x+e^x, & \text{if $x < 0$} \\
\ \alpha(1+x)^{2009}+\beta e^{-x}, & \text{if $x \ge 0$}
\end{cases} $$
I need to evaluate $\alpha$ and $\beta$, so that the function is continuously differentiable.

These are my thoughts:
Continuously differentiable $\Longleftrightarrow f^\prime(x) \ \text{exists and is continuous itself.}$ Thus we find the derivative of $f(x)$:
$$
f^\prime(x) = \begin{cases}
-\sin x+e^x, & \text{if $x < 0$} \\
\ 2009\alpha(1+x)^{2008}-\beta e^{-x}, & \text{if $x \ge 0$}
\end{cases}
$$
$\ f^\prime(x)$ definitely exists and is continuous, as:


*

*$-\sin x+e^x \ $ is continuous as a sum of two continuous functions

*$2009\alpha(1+x)^{2008}-\beta e^{-x} \ $ is continuous for the same reason

*now we need to pay special attention to $x=0$. $\ f^\prime(x)$ is continuous at $x=0$ $\Longleftrightarrow \lim_{x \to 0^-}-\sin x+e^x=2009\alpha(1+0)^{2008}-\beta e^{-0} \\$


From which we get $$2009\alpha - \beta = 1$$
However we got an expression for $\alpha$ and $\beta$ we still miss one equation in order to be able to determine both values. I am puzzled by the fact I don't know, where I can get the second expression from. What obvious thing am I missing? 
 A: As others have pointed out, you need to add that $f$ has to be continuos in $0$ as well.
You seem confused because wikipedia writes that you just need to check that $f'$ is continuos, and rightly so; in fact the fact that $f'$ exists implies that $f$ is continuos.
You may wonder then why you need to add a second equation; setting $f'$ continuos should result in $f$ being continuos automatically.
Basically, the point is that the expression you get for the derivative is not valid always, is valid only if $f$ is continuos. So you need to impose continuity of $f$, then your formula for $f'$ makes sense, and then you impose continuity of $f'$
A: You need $$\alpha + \beta = 2$$
in order for the function to be continuous at $x= 0$, a requirement for it to be continuously differentiable. Solving the system of this equation along with the one you have come up with simultaneously for $(\alpha,\beta)$ will give you the proper constants. Note that $$\lim_{x \to 0} \left[\cos(x) + e^x\right] = 2$$
and
$$\lim_{x \to 0} \left[\alpha(1 + x)^{2009} + \beta e^{-x}\right] = \alpha + \beta \,\,.$$
