f(x) = √x = x^{1/2}
f'(x) = 1/2*x^{-1/2} = 1/(2*√x)
mittlere Steigung im Intervall [0 ; 4]
m = (f(4) - f(0)) / (4 - 0) = (2 - 0) / (4 - 0) = 2/4 = 1/2
f'(x) = 1/(2*√x) = 1/2 --> x = 1
t(x) = f'(1) * (x - 1) + f(1) = 1/2 * (x - 1) + 1 = 1/2 * x + 1/2
Skizze
~plot~ sqrt(x);1/2*x;1/2*x+1/2 ~plot~