1. ## Tangent lines!!

alright...im trying to find the tangent line of

f(x) = ln(x) at x = 4

so i know that the slope simply equals f(x) which is

f(x) = 1/x = 1/4 = m

then i use that in the equation of a tangent line,

y = mx + b

then i use the values of x and y from f(x) = ln(x) and i get

f(4) = ln(4) = .4 so x = 4 , y = 1.4

then plugging these in the equation i get

1.4 = 1/4(4) + b and then this equals b = 1.4 - 1 = .4 so b = .4

now i plug all these values in and get

y = 1/4x + .4

but i got it wrong...can someone please tell me why and show me how i messed up...thanks

2. Originally Posted by ohmygodshoes
alright...im trying to find the tangent line of

f(x) = ln(x) at x = 4

so i know that the slope simply equals f(x) which is

f(x) = 1/x = 1/4 = m

then i use that in the equation of a tangent line,

y = mx + b

then i use the values of x and y from f(x) = ln(x) and i get

f(4) = ln(4) = .4 so x = 4 , y = 1.4

then plugging these in the equation i get

1.4 = 1/4(4) + b and then this equals b = 1.4 - 1 = .4 so b = .4

now i plug all these values in and get

y = 1/4x + .4

but i got it wrong...can someone please tell me why and show me how i messed up...thanks
The tangent point has the coordinates T(4, ln(4)).

The tangent has the slope $m = \frac14$

Now use point-slope-formula of the equation of a line:

$y-\ln(4) = \frac14 \cdot (x-4)$ . Solve for y:

$y = \frac14 x - 1 +\ln(4)$