# derivative and equation of tangent line of function with base e

• Feb 23rd 2009, 02:41 PM
williamb
derivative and equation of tangent line of function with base e
given f(x)= 4e^x -1x + 4

a) find derivative
I got 4ex -1 (which is correct)

***b) find the equation of a tangent line:

I thought that to find the equation of the tangent line you go
[f(x) - f(x+h)] / h
so i did this, plugged in
[(4ex -1 ) - (4e(x+h) -1 )] / h
cancelled and retrieved -4e for my answer.
which is wrong..

help plzz :(
• Feb 23rd 2009, 02:53 PM
skeeter
Quote:

Originally Posted by williamb
given f(x)= 4e^x -1x + 4

a) find derivative
I got f'(x) = 4e^x -1 (which is correct)

***b) find the equation of a tangent line:

general form for a tangent line to a curve at the point $(a, f(a))$

$y - f(a) = m(x - a)$

where $m = f'(a)$
• Feb 23rd 2009, 03:17 PM
williamb
where could you get the values from?
can you explain in more detail?
• Feb 23rd 2009, 03:35 PM
williamb
y-f(a) = m(x-a)

so m is f'(a) as in 4ex-1
and f(a) is 4e^x - x + 4 ?

y - 4e^x - x +4 = (4ex-1)(x-a)

??
is this right?
if so, what is a?
• Feb 23rd 2009, 04:10 PM
skeeter
$f(x) = 4e^x - x + 4$

$f'(x) = 4e^x - 1$

the tangent line to the curve $f(x) = 4e^x - x + 4$ at $x = a$ is ...

$y - f(a) = f'(a) \cdot (x - a)$

$y - (4e^a - a + 4) = (4e^a - 1) \cdot (x - a)$

normally, one is given a specific value of $x$ ... I'm using $x = a$ because it seems you were not given an $x$.
• Feb 23rd 2009, 04:49 PM
williamb
THANKS SO MUCH :D :D AWESOMEEEEE
you rock my world