Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
2  
It was a mistake. I corrected it. –  David Dec 29 '11 at 0:15

2 Answers 2

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.