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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?

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Find the equation that the differentiability condition gives. Then you have two equations in two unknowns... – David Mitra Dec 29 '11 at 0:00
If you want $f_L$ to be the function to the left of $1$ and $f_R$ to be the function to the right, you need to switch them around. – Alex Becker Dec 29 '11 at 0:12
It was a mistake. I corrected it. – Dave Dec 29 '11 at 0:15
up vote 1 down vote accepted

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.

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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 $a=20/3,b=-5/3$

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