Question

Find the linear approximation of f(x)=lnx at x=1 and use it to estimate ln1.38.
L(x)= ?
ln1.38 approximately = ?

Answer 1

the equation of the tangent to the curve is the approx

Answer 2

But I don't know how to do that, can you please help?

Answer 3

which part, finding a derivative, or working that into a linear equation?

Answer 4

Linear equation

Answer 5

y = mx - mPx + Py is a good rule to follow; this is where m=slope, and we know a point (Px,Py)

Answer 6

the derivative IS the slope of the curve ;

y = f'(x)x - f'(x) Px + Py

Px is simply the value ofx in the point used, in this case 1

y = f'(1)x -f'(1)(1) + Py

and Py is the value of the function at x=1; or simply .. f(1)


y = f'(1)x -f'(1) + f(1)

Answer 7

what is the derivate of ln(x) ?

Answer 8

1/x

Answer 9

By y do you mean f(x)?

Answer 10

I mean, by y do you mean L(x)?

Answer 11

yes, the linear function is y = L(x)

Answer 12

1/x at x=1 is 1

Answer 13

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

f'(1) = 1


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

and f(1) = ln(1) = 0

L(x) = x - 1 + 0

L(x) = x - 1 is the linear equation approximation

Answer 14

seems a little off .....

Answer 15

i see it; ln(1) = 0; 1-1=0 ... its good

Answer 16

ln(1.38) = abt 1.38-1 = .38

Answer 17

THANK YOU!!! :)

Answer 18

yep, hope it helps :)