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I am suppose to find the linearization of a function at a, I don't know what an a is.

$f(x) = x^4 + 3x^2, a= -1$ is an a suppose to be like a y?

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$a$ is just a letter. It has no special significance, but can stand for a number. Here it stands for $-1$ (so the problem tells you), so the task is to find the linearization of the function at $-1$. In this context, "at" means that $-1$ is the argument to the function, that is, the $x$ in $x^2+3x^2$. – Henning Makholm Oct 4 '11 at 2:10
So why don't they just say x = -1? – user138246 Oct 4 '11 at 2:12
It just feels like something is missing from this. I have L(x) = fprime(a)(x-a) +f(a) so I know that the problem should be $4x^3 + 6x (x-a) +2 $ but something is missing still. – user138246 Oct 4 '11 at 2:14
You had $L(x) = f'(a) \cdot (x-a) + f(a)$ which is correct. The derivative computation $f'(x) = 4x^3+6x$ is also correct. But in the formula for "linearization", you need to plug in $f'(a)$, not $f'(x)$. Remember that finally $L(x)$ should be a linear function when considered as a function of $x$. (It will/need not be linear in $a$.) – Srivatsan Oct 4 '11 at 2:45
They could have said ‘at $x=-1$’, but they wanted to be careful to distinguish $x$, the variable, from $a$, a particular real number. Then they told you which real number $a$ is: it’s $-1$. Since it has a known value, there will be no letter $a$ anywhere in your work for this problem. – Brian M. Scott Oct 4 '11 at 2:48
up vote 2 down vote accepted

It's asking you to find the tangent line to its graph when $x=-1$. To do this, you need to take it's derivative and find $f'(-1)$. Then point-slope form for a line will give you the tangent line.

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f(x)=a? Or am I wrong? – user138246 Oct 4 '11 at 2:00
@Jordan: No, $f(x)\ne a$; $f(x)=x^4+3x^2$. The linearization of $f$ at $a$ is, as Joe said, the tangent line to the graph of $y=x^4+3x^2$ at $x=a$. Here $a=-1$, so you’re looking for the tangent line at the point $(-1,4)$. – Brian M. Scott Oct 4 '11 at 2:43
So x = -1 and y = f(x) – user138246 Oct 4 '11 at 2:47
@Jordan: Sort of, but not really. In the formula $y=f(x)$, $x$ is a variable: it has no fixed value. However, you’re calculating the linearization of $f$ at a particular value of $x$, namely at $x=-1$. You could do the same thing at the particular value $x=a$, getting $L(x)=f'(a)(x-a)+f(a)$, and then simply substitute $-1$ for $a$ throughout; that’s what Srivatsan is explaining up above. – Brian M. Scott Oct 4 '11 at 2:54
So I am calculating the problem a a and a is -1 and a= -1 and x= -1? So then I have fprimeof(-1)(1+1) + f(a) which is 4(-1^3) + 6(-1) (-1--1) +2? – user138246 Oct 4 '11 at 3:03

One way to think about differentiable functions is they can locally be well-approximated by lines. Here is the function $f(x) = x^4 + 3x^2$ and the linearization (which I'll leave up to you to find):

enter image description here

The idea is that the tangent line at the point $(a,f(a))$ will well approximate the function $f(x)$ for $x$-values "near" $a$. As you can see, the line really does approximate the function for $x$-values close to $-1$. The closer the $x$-value is to $-1$, the better the line will approximate the function. To find the linearization of a function $f(x)$ at $x = a$, there is a fancy formula in Stewart's (and to be fair, most other texts on the subject of) calculus. However, this formula really just wants you to find the tangent line of the function $f(x)$ at the point $(a,f(a))$. The slope of this tangent line is given by $f'(a)$ by the definition of the derivative, and the slope and point $(a, f(a))$ are enough to give you the equation of the line.

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Well it is beyond frustrating to be working on this single problem for the better part of a day. – user138246 Oct 4 '11 at 3:17
Frustration is a different story, and it is normal. If a problem causes you to get this upset, I would suggest putting down the problem, and revisiting it at another time. – JavaMan Oct 4 '11 at 3:19
I don't have that kind of time and I attempt to. I just wish people like Stewart weren't so arrogant about math and tried to explain things in a way that people who aren't amazing at math can learn from. – user138246 Oct 4 '11 at 3:24

The "linearization of $f(x)$" means "the linear function which best approximates $f(x)$." The graph of a linear function is a line, while $f(x)$ can look pretty weird. So no linear function is going to look like $f$ everywhere. But if we choose some point on the graph of $f(x)$ -- that is, if we look at "the linearization of $f(x)$ at -1" -- we can get a good linear approximation of $f(x)$ near that point.

The thing is, the linearization of $f(x)$ is also a function, which you've written $L(x)$. It might get confusing to talk about "the linearization of f(x) at x = -1" because both $L$ and $f$ use $x$ as a variable. So we usually talk about the linearization at $a$, which is a perfectly fine letter.

You start with $f(x) = x^4 + 3x^2$, and you want to find its linearization at $a = 1$. You already have a formula for it: $L(x) = f'(a)(x-a)+f(a)$.

$f'(a) = 4a^3 + 6a$ and $f'(-1) = -4 - 6 = -10$. So $L(x) = -10(x - (-1)) + f(-1) = -10(x+1) + 4$.

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I don't follow those last two lines at all. – user138246 Oct 4 '11 at 2:57
The derivative of $f$ is a function. I assume you know how to calculate $f'(x) = 4x^3 + 6x$. Well, if we want to calculate the value of $f'(x)$ when $x = -1$, we calculate $f(-1) = -4 - 6 = -10$. – Adam Saltz Oct 4 '11 at 3:04
Uh, $(-1)^4 = (-1)\times(-1)\times(-1)\times(-1) = 1$. – Adam Saltz Oct 4 '11 at 3:27
Your difficulties might lie more with algebra than calculus. You ought to be able to compute $(-1)^4 = 1$ by hand or in your head. In any case, you're either inputting the wrong thing into your calculator or your calculator is wrong. – Adam Saltz Oct 4 '11 at 3:32
@Jordan "I think my calculator assumes $−1^2$ is going to be -(1^2) for some reaosn." - Most people follow exactly the same convention the calculator follows. If you want to specify $(-1)^4$, you should clearly put the paranthesis around $-1$. – Srivatsan Oct 4 '11 at 4:23

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