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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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Notation: does $f^1(a)$ mean $f'(a)$? –  Américo Tavares Sep 11 '11 at 21:05
    
If you want to find $f'(a)$, you compute $f'(x)$ and then substitute $x=a$. –  Américo Tavares Sep 11 '11 at 21:07
    
Yes $f'(a)$ is what I meant.I dont know what but the f(a) to f(x) and f(a+h) stuff is really confusing to me, I thought I understood how it worked but I forgot. –  user138246 Sep 11 '11 at 21:09
    
Do are still confused with $f(a)$ and $f(a+h)$ in this specific case? –  Américo Tavares Sep 11 '11 at 21:55
    
Not as much, I get what to do but it still confused me a little. I mean I can plug in numbers now but I don't quite understand why. –  user138246 Sep 11 '11 at 21:58

2 Answers 2

up vote 2 down vote accepted

$f(a)$ 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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I got -6a-4 which is wrong, it should be 6a-4. Also here is where I demonstrate how terrible at math I am. Isnt 3a x 3h going to be 9ah? –  user138246 Sep 11 '11 at 21:19
    
@Jordan: That's because I put in the wrong sign in $6a$. I'll fix it. Okay, fixed. You should be getting the right answer now. And while it is true that $3a\times 3h=9ah$, this does not appear in the expression. Note that $3(a+h)^2$ is not the same thing as $(3a+3h)^2$. First do the square, then multiply by $3$. –  Arturo Magidin Sep 11 '11 at 21:23
    
I think that it's better I delete my answer, since you have posted a better one in the meantime. –  Américo Tavares Sep 11 '11 at 21:36
    
@Américo: I was writing the original while you were; I think you just posted first. –  Arturo Magidin Sep 11 '11 at 21:38

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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Judging from previous questions, the OP is probably still at the stage of developing these rules. –  Arturo Magidin Sep 11 '11 at 21:16
    
I have no idea what just happened. Where did $f'(x)=3\times 2x-4=6x-4$ come from? –  user138246 Sep 11 '11 at 21:16
    
@Arturo Magidin, Jordan Carlyon: I will add the computation from the definition. –  Américo Tavares Sep 11 '11 at 21:22
    
@Jordan: Américo is asking if you know "the basic rules for computing the derivative of a function" (i.e., the power rule, the product rule, etc) but if you don't, then this will not make sense (yet). –  Arturo Magidin Sep 11 '11 at 21:22
    
Are you talking about the limit laws? –  user138246 Sep 11 '11 at 21:32

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