Function parameters and continuity Let $$f(x)=\begin{cases}
\frac{a\log x}{5(x-1)}&;x>1\\ \\
-\frac{bx^2}5+x&;x\leq 1.
\end{cases}$$
I would like to find $a$ and $b$, so that the function will be continuous and have a derivative at 1.
Lets say that $f_{R}(x) = \frac{a\log(x)}{5(x-1)}$ and $f_{L}(x) = -\frac{bx^2}5+x$.
The first thing I did was to find the limit of $f_{R} = \frac{a}{5}$ and $f_{L}=\frac{5-b}{5}$. The function will be continuous, if these two limits will be equal.
How would I proceed and how would I find a and b so that it would fit the criteria at the beginning of this post?
 A: so from what you have, you get $a=5-b$.  to get a second condition note that the derivative of $f$ on $(-\infty,1)$ is $1-2bx/5$, while on $(1,\infty)$ it is
$$\frac{a(1-1/x-\log x)}{5(x-1)^2}$$  the derivatives will match at $x=1$ if (taking limits as $x\to1$ from the left and right)
$$1-2b/5=-a/10$$
so we have the two conditions
$$
a+b=5, a+4b=0
$$
which give $a=20/3,b=-5/3$
A: I will assume you mean $f_{R}(x) = \frac{a\log(x)}{5(x-1)}$ and $f_{L}(x) = -\frac{bx^2}5+x$ so that these correspond to right and left. You have the condition for continuity already, that $\frac{a}{5}=\frac{b-5}{5}$. Recall that a function $f(x)$ is differentiable at $1$ if and only if the limit
$$\lim\limits_{h\to 0} \frac{f(1+h)-f(1)}{h}$$ exists, and for the limit to exist you need the right and left limits to be equal, i.e.
$$\lim\limits_{h\to 0-} \frac{f(1+h)-f(1)}{h} = \lim\limits_{h\to 0+} \frac{f(1+h)-f(1)}{h}$$ which in your case is the condition 
$$\lim\limits_{h\to 0-} \frac{f_L(1+h)-f_L(1)}{h} = \lim\limits_{h\to 0+} \frac{f_R(1+h)-f_L(1)}{1}$$ 
or after substituting
$$\lim\limits_{h\to 0-} \frac{-\frac{b(1+h)^2}{5}+\frac{b}{5}}{h} = \lim\limits_{h\to 0+} \frac{\frac{a\log(1+h)}{5h}+\frac{b}{5}}{h}$$ 
which I will allow you to evaluate. This will give you a second equation that $a$ and $b$ must satisfy, so you can find $a$ and $b$ by solving the resulting system of two equations.
