# Thread: Linear Aprox / differentiatials.

1. ## Linear Aprox / differentiatials.

Looking back on this problem, i have one question..
Its wanting to use a linear approximation to estimate the number..
(2.001)^5

Here's my work: x=2..
f(x)= x^5 f(2) = 32 = (which is y),
f'(x)= 5x^4. f'(2) = 80 = (which is m)

Y-32=80(x-2)
y=80x-128

so i know (2.001)^5 actually is 32.08008004...
and y=80(2)-128 = 32...
and y=80(.001) - 128 = .08 - 127.92 = -127.92...

then the answers were that the ▲y = .08008 and dy=.08
what are ▲y=.08008 and dy from?
If the .08 is from y=80(.001), where did the -128 go?
thanks...

2. Originally Posted by 3deltat
Looking back on this problem, i have one question..
Its wanting to use a linear approximation to estimate the number..
(2.001)^5

Here's my work: x=2..
f(x)= x^5 f(2) = 32 = (which is y),
f'(x)= 5x^4. f'(2) = 80 = (which is m)

Y-32=80(x-2)
y=80x-128

so i know (2.001)^5 actually is 32.08008004...
and y=80(2)-128 = 32...
and y=80(.001) - 128 = .08 - 127.92 = -127.92...

then the answers were that the ▲y = .08008 and dy=.08
what are ▲y=.08008 and dy from?
If the .08 is from y=80(.001), where did the -128 go?
thanks...
there are two basic ways to approach a problem like this:

the one i learned is to use the linear approximation equation:

$f(x) \approx f(a) + f'(a)(x - a)$

here, x is the value that you want to evaluate the function at, while a is a value close to x that you already know the value of the function for

the other way is one i picked up while tutoring.

$f(x + dx) = f(x) + dy$ ....er, or something like that... i know how to use it, that's what's important .

anyway, here x is the value you can evaluate the function at, dx is a small change to this value, f(x) is the value of f at x and $dy = f'(x) \cdot dx$ (we get this by solving $\frac {dy}{dx} = f'(x)$ for $dy$)

which method do you want to use?

3. The f(x)=f(a) + f'(a)(x-a) is one i've learned before aye. For some reason the teacher didnt like it and went to the slope formula directly.

still, it'd come up with y=80x -128 no?

4. Originally Posted by 3deltat
The f(x)=f'(a)(x-a) is one i've learned before aye. For some reason the teacher didnt like it and went to the slope formula directly.

still, it'd come up with y=80x -128 no?
apparently you have a different method from mine.

my method:

define $f(x) = x^5$

then $(2.001)^5 = f(2.001) \approx 2^5 + f'(2)(2.001 - 2)$

now just work out the right hand side to get the answer. which would be 32.08