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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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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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):
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.
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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