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Where am I going wrong?

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

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

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

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

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

1 Answers 1

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Well, nowhere. Everything is OK.

  • 0
    But shouldn't the answer be -6x-1?2012-05-30
  • 0
    @AFerraro: That is when $h \to 0$ which happens to be the derivative :)2012-05-30
  • 0
    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$2012-05-30
  • 1
    @AFerrara It will be $-6x-1$ only when you take the limit as $h \rightarrow 0$.2012-05-30
  • 0
    Ok, I thought they were the same, Thanks2012-05-30