Question

the linearization of f(x) = 5x^2 + 3x -2 near x =1 is (show work please)

Answer 1

In general, given a function f which is differentiable at a point p, the linearization L of f at p is given by

L(x-p) = f(p) + f'(p)(x-p)

So using that formula, find L for your f and p.

Answer 2

Actually for you, you'll want to use

L(x) = f(p) + f'(p)(x-p) ----- (*)

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The motivation for this formula is hopefully clear: the tangent line at p is the best linear approximation odor the graph of f. This equation (*) for a straight line is a the equation of that tangent line to the graph of f at that point.

Answer 3

could you please explain the steps to solve, a little confused as to what variable represents what.

Answer 4

For example, let's linearize f(x) = x^2 at x = 1.
First f'(x) = 2x thus f'(1) = 2. Also f(1) = 1.

Hence the linearization is

L(x) = f(1) + f'(1)(x-1)
= 1 + 2(x-1)
= 2x - 1

This is the equation of the tangent line of y = x^2 at x = 1.
For x = 1, L(1) = 1 = f(1), which is exactly what we'd expect.

Answer 5

I.e., the linearization of f(x) = x^2 at p = 1 (or x = 1) is
L(x) = f(p) + f'(p)(x-p)
= 2x - 1

Answer 6

ok got it trying it out thanks

Answer 7

Answer : y = 13x - 7

Answer 8

Let's see.

If
f(x) = 5x^2 + 3x -2, p = 1

then
f'(x) = 10x + 3.
f(p=1) = 6, f'(1) = 13

hence

L(x) = 6 + 13(x-1)
= 13x - 7

Yep.

Answer 9

thanks