prove-that-the-function-f-x-1-e-x-2-2x-has-a-root-in-the-open-interval-0-1-recall-that-e-2-7-

Question Number 125815 by Tanuidesire last updated on 14/Dec/20

p r o v e t h a t t h e f u n c t i o n f (x) = \frac{1}{e^{x} - 2} - 2 x, h a s a r o o t i n t h e o p e n i n t e r v a l (0, 1) . r e c a l l t h a t e \approx 2.7

Answered by physicstutes last updated on 14/Dec/20

f (x) = \frac{1}{e^{x} - 2} - 2 x

f (0) = \frac{1}{- 1} - 0 = - 1

f (1) = \frac{1}{e - 2} - 2 \approx - 0.6

f (1) \times f (0) > 0 hence f (x) does not have roots in that inteva

Commented by Tanuidesire last updated on 14/Dec/20

t h a n k s!

Commented by mindispower last updated on 14/Dec/20

f (0) . f (1) < 0 w i t h e c o n t n u i t y \Rightarrow f (x) = 0 h a s e s o l u t i o n

f (0) . f (1) > 0 w e c a n s a y n o t h i n g

e x e m p l

f (x) = x + 1, f (0) . f (1) = 2 > 0 f (x) = 0 h a s e n o s o l u t i o n

f (x) = {(x - \frac{1}{2})}^{2}

f (0) f (1) = \frac{1}{16} > 0 f (x) = 0 h a s e s o l u t i o n x = \frac{1}{2}