If $f(a) < 0$ and $f(b) > 0$, then prove that the point $c$ computed in the bisection method is the point where the line through $(a, \operatorname{sign}(f(a)))$ and $(b, \operatorname{sign(f(b))})$ intersects the $x$-axis.
Can anyone give me hints on this homework question of mine?
$(a, sign(f(a)) = (a, -1)$, $(b, sign(f(b)) = (b, +1)$. This is because $f(a) < 0$, hence $sign(f(a)) = -1$ and vice versa for $sign(f(b))$.
Now, the straight line between $(a, -1)$ and $(b, +1)$ has slope
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{+1 - -1}{b - a} = \frac{2}{b - a} $$
Hence, the general equation of the line is $$ y - y_1 = m(x - x_1) \\ y - 1 = \frac{2}{b - a}(x - b) $$
The $x$ intercept is found where $y = 0$.
Plugging that into the equation yields
$$ 0 - 1 = \frac{2}{b - a}(x - b) \\ -1 (b - a) = 2(x - b) \\ a - b = 2x - 2b \\ 2x = a + b \\ x = \frac{a + b}{2} $$
Which is the exact point that the bisection method yields (the point with $x= \frac{a + b}{2}$)
