Question

suppose that $f:[0,1]\rightarrow r$ is continous on [0,1] and differentiable on (0,1) with f(0)=0 anf f(1)=10.prove that there exist 100 distinct points x_k belonging to (0,1) such that summation from k=1 to 100 of 1/f'(x k)=1000

Answer 1

equation of normal to f(x) at point (x_k, y_k):

$y-y_k=-(x-x_k)/f'(x_k)$

so,

$\frac{1}{f(x_k)}=\frac{y-y_k}{x_k-x}$

then for each x_k we can take values of x and y such that

$\displaystyle\sum_{k=1}^{100}\frac{1}{f(x_k)}=\displaystyle\sum_{k=1}^{100}\frac{y-y_k}{x_k-x}=1000$