Linear approximations

[fk378]

1. The problem statement, all variables and given/known data
explain in terms of linear approximations why the approximation is reasonable.
(1.01)^6=1.06



2. Relevant equations
y-y1=m(x-x1)
L(x)=f(a) + f'(a)(x-a)


3. The attempt at a solution
given 1.01^6, f(x)=x^6, so f'(x)=6x^5
plugging in x=1, f'(1)=6
y-6=6(x-1)

Is that the right equation?
Because in class the equation was
y-1=6(x-1)
y=1.06
which is the correct answer. But why is y1=1 and not 6??? Can anyone catch that?

[Feldoh]

which is the correct answer. But why is y1=1 and not 6??? Can anyone catch that?

y-y1=m(x-x1)

Where m is your slope and the point given is (x1,y1), since f(1)=1 your point is (1,1)
f'(1)=6=m so...

y-1=6(x-1)

[fk378]

They didn't give me P(1,1) though
I arbitrarily picked x=1 to solve the original f'(x) equation.

[fk378]

y-y1=m(x-x1)

Where m is your slope and the point given is (x1,y1), since f(1)=1 your point is (1,1)
f'(1)=6=m so...

y-1=6(x-1)

Isn't my point (1,6)
I plugged in x=1 and got f'(x)=6

[Feldoh]

Isn't my point (1,6)
I plugged in x=1 and got f'(x)=6

Yes f'(1)=6 but that doesn't mean when x=1 f(x)=6 necessarily. The point of a function f(x) at x is (x,f(x)). What does the derivative tell us about f(x)?

[fk378]

So f'(x)=6, but f(x)=1 so f(x) is the value I should put in for y in (x1,y1)?

[HallsofIvy]

They didn't give me P(1,1) though
I arbitrarily picked x=1 to solve the original f'(x) equation.
I hope it was NOT "arbitrary". The fact that 1 is very close to 1.01 should have guided you!

So f'(x)=6, but f(x)=1 so f(x) is the value I should put in for y in (x1,y1)?
f(x) is NOT "1", the function you are looking at is f(x)= x6. What is f(1)= (1)6?