I know that the tangent is on the form $y=ax+b$ but how should I show this one:
Let $f:\mathbb{R}→\mathbb{R}$ be a differentiable function. Consider a sample point $(x_0, f(x_0))$.
(a) determine the tangent line at $f(x_0)$, i.e. the straight line $l(x)=ax+b$, that passes through $f(x_0)$ with slope $a = f'(x_0)$.
(b) compute the intersection of $l$ with the $x$-axis, i.e. compute the root of $l$.
I am pretty sure I know how I can solve (b) if I know what (a) is.
What do we know about the tangent line at the point $(x_0, f(x_0))$? We know it has slope $f'(x_0)$ and it passes through the point $(x_0, f(x_0))$. Remember the point-slope form of the equation of a line? $y-y_0 = m(x-x_0).$
Alternatively, you could start with $y=mx+b$ and plug in the slope and the know point
$$f(x_0) = f'(x_0) x_0+b$$
and solve for $b$.
The tangent line has a common point $l(x_0) = f(x_0)$ and the slope in common with $f$: $l’(x) = f’(x_0)$.
Here $l’(x) = a$, and $l(x_0) = a x_0 + b = f(x_0)$.
This results in $l(x) = f’(x_0) (x-x_0) + f(x_0)$ which I remember by the first order Taylor expansion of $f$ around $x_0$.
