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I am suppose to find f'(a) for $f(x) = 3x^2 - 4x + 1$

I am having trouble because I don't know what $f(a)$ is equal to and I can't seem to figure out how to approach it from this way.

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    See [this question](http://math.stackexchange.com/questions/61033/finding-limit-of-a-quotient) of yours too.2011-09-11

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$f(a)$ is equal to $f(a) = 3a^2 - 4a + 1$. The $a$ is left indicated, because it is only given "indicated" to you, not as an actual value.

The derivative is equal to the limit: \begin{align*} f'(a) &= \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ &= \lim_{h\to 0}\frac{ \bigl(3(a+h)^2 -4(a+h) + 1\bigr) - \bigl(3a^2-4a+1\bigr)}{h}\\ &=\lim_{h\to 0}\frac{\bigl( 3(a^2+2ah+h^2) - 4a-4h +1\bigr) - \bigl(3a^2-4a+1\bigr)}{h}\\ &=\lim_{h\to 0}\frac{3a^2 + 6ah + 3h^2 - 4a - 4h + 1 - 3a^2 + 4a - 1}{h}\\ &=\lim_{h\to 0}\frac{3h^2 + 6ah - 4h}{h}. \end{align*} Up to here, all we've done is use the definition of $f$, and do algebra on the expression in the limit, nothing else.

At this point: if you try plugging in $h=0$ into the limit, you get $\frac{0}{0}$, which is what you should expect (computing limits of difference quotients always give $\frac{0}{0}$ at first). But there is clearly a factor of $h$ to be factored out of the numerator; factor it, cancel it with the denominator, and do the resulting limit. The answer should be an expression that involves $a$ but not $h$ (since the question is in terms of $a$, the answer will be in terms of $a$ as well).

(The whole point here is to realize that for any particular value of $a$, the computations of the limit, and so of the derivative, are actually the same: substitue $a=1$ and you can do exactly the same steps as above to get the value f'(1); substitute $a=2$, and again the exact same steps work to find f'(2); substitute $a=\pi$, and the same steps work to compute f'(\pi); substitute $a=1058431278903210532.5789432\sqrt{2}$, and the same steps work to compute f'(1058431278903210532.5789432\sqrt{2}). So instead of doing all the work each time we need the value of the derivative at a point, we can just do the work once, and get an answer into which we will just need to "plug in" whatever number we need to get the answer.)

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    @Américo: I was writing the original while you were; I think you just posted first.2011-09-11
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Do you know the basic rules for computing the derivative of a function?

Assuming so, for $f(x)=3x^2-4x+1$, you have f'(x)=3\times 2x-4=6x-4. Then f'(a)=6a-4.

If you have to compute f'(a) from the definition, evaluate

$f^{\prime }(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}.$

Added in response to OP's comment.

You have $f(x)=3x^{2}-4x+1$. So $f(a)=3a^{2}-4a+1$ and

$f(a+h)=3\left( a+h\right) ^{2}-4\left( a+h\right) +1.$

Thus$^1$ $\begin{eqnarray*} f^{\prime }(a) &=&\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} \\ &=&\lim_{h\rightarrow 0}\frac{3\left( a+h\right) ^{2}-4\left( a+h\right) +1-\left( 3a^{2}-4a+1\right) }{h} \\ &=&\lim_{h\rightarrow 0}\frac{6ah+3h^{2}-4h}{h}\quad \text{you have to simplify this fraction as shown below} \\ &=&\lim_{h\rightarrow 0}\; 6a+3h-4 \\ &=&6a-4. \end{eqnarray*}$

-- $^1$ Detailed computation. From $\begin{eqnarray*} 3\left( a+h\right) ^{2} &=&3a^{2}+6ah+3h^{2} \\ -4\left( a+h\right) +1 &=&-4a-4h+1 \\ -\left( 3a^{2}-4a+1\right) &=&-3a^{2}+4a-1 \end{eqnarray*}$

we get $\begin{eqnarray*} &&3\left( a+h\right) ^{2}-4\left( a+h\right) +1-\left( 3a^{2}-4a+1\right) \\ &=&3a^{2}+6ah+3h^{2}-4a-4h+1-3a^{2}+4a-1=6ah+3h^{2}-4h \end{eqnarray*}$

and for $h\ne 0$

$\frac{6ah+3h^{2}-4h}{h}=\frac{h(6a+3h-4)}{h}=6a+3h-4$

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    I don't think so.2011-09-11