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I am having problems with this question, it would be wonderful if someone can help.

Given that $f(x)= x^2 + x - 3$

1) Find $f(x + h)$

2) Then express $f(x+h)-f(x)$ in its simplest form

3) Deduce $\lim\limits_{h->0}\dfrac{f(x+h)-f(x)}{h}$

Thanks for the help, i was stuck on the second part.

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    $\begingroup$ have you done the first 2 parts? where are you stuck? $\endgroup$ – Kevin Nov 30 '13 at 19:18
  • $\begingroup$ What have you done so far? Where are you having problems, specifically? $\endgroup$ – Strants Nov 30 '13 at 19:18
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  1. Just substitute: $f(x+h) = (x+h)^2 + (x+h) - 3$

    • $x^2 + 2xh + h^2 + x + h - 3$
  2. Substitute again $f(x+h) - f(x) = \left((x+h)^2 + (x+h) - 3\right) - (x^2 +x - 3)$

    • Then simplify!

    • $(x^2 + 2xh + h^2 + x + h - 3) - (x^2 +x - 3)$

    • $x^2 + 2xh + h^2 + x + h - 3 - x^2 - x + 3$

    • $2xh + h^2 + h$

  3. Hint: Stick your answer to 2 in over $h$ and see what you can do.

    • $\lim\limits_{h->0}\dfrac{2xh + h^2 + h}{h}$

    • $\lim\limits_{h->0}(2x + h + 1)$

    • $2x + 1$

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If $f(x) = x^2 + x - 3$ and $h$ is a real number, then $f(x + h) = (x + h)^2 + (x + h) - 3$. After some simplification, we find that $f(x + h) - f(x) = h(2x + h + 1)$. Taking the quotient of $f(x + h) - f(x)$ and $h$ and letting $h \rightarrow 0$, we find by substitution that $f'(x) = 2x + 1$.

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