How do you find the formula for a tangent line if there are no points?

Question

The question asks what is the function for the tangent line $b(x)$ if

$$f(x) = \frac{e^{x}}{e^{x}+1}$$

and the part that gets me is the point given is $P = (a,f(a))$.

I derived the gradient and used the slope point formula and subbed it in but to no avail.

Any hints?

here is the work so far ill shorten some of it:

Point slope formula $y-f(a) = m(x-a)$

$m = \frac{e^a}{(e^a+1)^2}$ (first derivative of f(x))

$y-f(a) = \frac{e^a}{(e^a+1)^2}(x-a)$

$y = \frac{e^x}{e^x+1}+\frac{e^a}{(e^a+1)^2}(x-a)$

this is where I am stuck for $(x-a)$ if I sub it in it's wrong so is $a =f(x)$ like $y$?

Answer 1

You say "there are no points" but (a,f(a)) is a point! If $f(x)= \frac{e^x}{e^x+ 1}$ then $f'(x)= \frac{e^x(e^x+ 1)- e^{2x}}{(e^x+ 1)^3}= \frac{1}{e^{2x}+ 2e^x+ 1}$. At x= a that is $\frac{1}{e^{2a}+ e^a+ 1}$. Of course, $f(a)= \frac{e^a}{e^a+ 1}$ so the tangent line at (a,f(a)) is

$y= \frac{1}{e^{2a}+ 2e^a+ 1}(x- a)+ a\frac{e^a}{e^a+ 1}$

Answer 2

Taylor series of 1st order

$$ b(x) = f(a) + (x-a) \lim_{x\rightarrow a} \tfrac{\rm d}{{\rm d}x} f(x) $$

Answer 3

You just made a slight error there, you put in $f(x)$ instead of $f(a)$, $f(a) = \frac{e^a}{e^a +1}$

So $b(x) = \frac{e^a}{(e^a +1)^2} (x-a)+ \frac{e^a}{e^a +1}$