The problem is a precalculus problem.

$$\frac{\large\frac{1}{1+x+h} - \frac{1}{1+x}}{h}$$

I was wondering if I can use the distributive property by dividing out the denominators in the numerator. My plan was to solve the fraction in the numerator then multiply by the reciprocal. However, since the denominator in the numerator is a sum, I am unsure as how to do it.

I know I could simplify / reduce the expression by dividing the polynomial by a monomial, but I am just confused as of the steps and concept. Can someone clarify?


Math student

  • 2
    $\begingroup$ It would be nice if you could use this guide and tidy up your expression, because as it stands, there's just too many possible interpretations of what terms are numerators and denominators to make any sense of it. $\endgroup$ – Arthur Mar 1 '14 at 15:07
  • 2
    $\begingroup$ At least, type the parentheses if required. $\endgroup$ – Claude Leibovici Mar 1 '14 at 15:08
  • 1
    $\begingroup$ The question has been edited, but clearly not to what OP wanted. I also cannot figure out what it is supposed to look like. $\endgroup$ – Ross Millikan Mar 1 '14 at 15:31
  • $\begingroup$ The h divides both rational expressions. $\endgroup$ – Cetshwayo Mar 1 '14 at 15:34
  • $\begingroup$ How's that? At least it looks like the difference quotient for $f(x)=1/(1+x)$. $\endgroup$ – Mark McClure Mar 1 '14 at 15:35

Here's how WolframAlpha says to do it. There's a few extra steps than Ross's answer (as you requested) but, otherwise, it's pretty much what he did. So anyone who upvotes this has to upvote his. I doubt I'd upvote either one, myself.

enter image description here

  • $\begingroup$ Thanks for this explanation. Thus, for a problem such as this it is better to treat the denominators as entire terms rather than sums, and multiplying them by their LCD? $$\frac{\large\frac{1}{1+x+h} - \frac{1}{1+x}} $\endgroup$ – Cetshwayo Mar 1 '14 at 16:03
  • $\begingroup$ @Utvecklaochförenkla I'm not sure what you mean by "treat the denominators as entire terms". Perhaps you mean that there's no reason to expand the $(1+x+h)(1+x)$ in the denominator, which is certainly true. $\endgroup$ – Mark McClure Mar 1 '14 at 16:05
  • $\begingroup$ Maybe I am using the wrong terminology but is not (1+x+h) a polynomial with three terms? Is (1+x+h) not considered a term when it is being subtracting by another rational expression? $\endgroup$ – Cetshwayo Mar 1 '14 at 16:11
  • $\begingroup$ I am really trying to have a more intuitive understanding as to how to solve these types of problems. Moreover, I am seeking to understand the "Property of Real Numbers" better. Nonetheless, when I see sums with numbers and variables in the denominator I always get confused as to whether I can multiply them by the LCD, or if that is not allowed. For example, can I consider (1+x+h) and (1+x) a term, such as b and d, respectively in a/b - cd = ad-cb / bd? $\endgroup$ – Cetshwayo Mar 1 '14 at 16:22
  • $\begingroup$ @Utv So is the essence of your question really about whether the laws of real numbers also apply to expressions composed of real numbers and variables, e.g. quotients of polynomials? If so, to answer that it would be helpful to know more about your level of study. Do you know calculus and/or abstract algebra? $\endgroup$ – Bill Dubuque Mar 1 '14 at 16:38

You have $$\frac{\large\frac{1}{1+x+h} - \frac{1}{x+1}}{h}=\frac 1h \left(\large\frac{1}{1+x+h} - \frac{1}{x+1}\right)\\=\large\frac{1}{h(1+x+h)} - \frac{1}{h(x+1)}\\=\frac{(x+1)-(x+1+h)}{h(x+1)(x+1+h)}\\=\frac {-h}{h(x+1)(x+1+h)}$$ and if $h \neq 0$ you can divide top and bottom by it to get the result you are looking for.

  • $\begingroup$ I do appreciate your help but you left out to many steps for me to follow how you solved. $\endgroup$ – Cetshwayo Mar 1 '14 at 15:49
  • $\begingroup$ I added one step at the start. For the second line, I distributed the $\frac 1h$. For the third, I put them over a common denominator by cross multiplying. $\endgroup$ – Ross Millikan Mar 1 '14 at 15:53
  • $\begingroup$ This is very probably an exercise towards showing that $\;\left(\frac1{1+x}\right)'=-\frac1{(1+x)^2}\;$ . At this level, I can't understand how the OP can't do or doesn't understand sum/substraction of fractions...+1 , of course. $\endgroup$ – DonAntonio Mar 1 '14 at 17:10

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