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.

Where am I going wrong?

Find the difference quotient for: $f(x)=2-x-3x^2$

$$\frac{[ 2-(x+h)-3(x+h)^2 ] - [ 2-x-3x^2 ]}{h}$$

$$\frac{2-x-h-3x^2-6hx-3h^2-2+x+3x^2}{h}$$

$$\frac{-3h^2-6hx-h}{h}$$

$$-3h-6x-1$$

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Well, nowhere. Everything is OK.

share|improve this answer
    
But shouldn't the answer be -6x-1? –  AFerrara May 30 '12 at 2:33
    
@AFerraro: That is when $h \to 0$ which happens to be the derivative :) –  user9413 May 30 '12 at 2:33
    
Yes, but you need to take the limit, and what you have found is the quotient. To find the derivative you have to let $h \to 0$ –  Pedro Tamaroff May 30 '12 at 2:34
1  
@AFerrara It will be $-6x-1$ only when you take the limit as $h \rightarrow 0$. –  user17762 May 30 '12 at 2:34
    
Ok, I thought they were the same, Thanks –  AFerrara May 30 '12 at 2:35

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.