Ja. Beides ist richtig. Prima gemacht.
a)
f(x) = 1/x - 1
f'(x) = -1/x^2
f(x)/x = f'(x) --> x = 2
t(x) = f'(2)·(x - 2) + f(2) = - 1/4·x
b)
f(x) = √x - 1
f'(x) = 1/(2·√x)
f(x)/x = f'(x) → x = 4
t(x) = f'(4)·(x - 4) + f(4) = 1/4·x
Skizze
~plot~ 1/x-1;sqrt(x)-1;-1/4*x;1/4*x;[[-1|7|-3|3]] ~plot~