elementary-calculus-if-f-x-x-3-x-2-2x-2-x-3-1-x-3-1-3-2-cos-pix-3-then-f-0-amp-f-1-

Question Number 141274 by mnjuly1970 last updated on 17/May/21

\dots \dots . e l e m e n t a r y \dots . . c a l c u l u s \dots \dots . .

i f f (x) := \frac{\sqrt[3]{(x^{3} + x - 2) (2 x^{2} + x - 3) (1 - x^{3})}}{{(2 + c o s (π x))}^{3}}

t h e n f^{'} (0) = ? & f^{'} (1) = ?

Answered by mindispower last updated on 17/May/21

l n (f (x)) = \frac{1}{3} (l n (2 - (x^{3} + x)) + l n (3 - 2 x^{2} - x) + l n (1 - x^{3})) - 3 l n (2 + c o s (π x))

\frac{f^{'} (x)}{f (x)} = \frac{1}{3} (\frac{- 3 x^{2} - 1}{2 - x^{3} - x} - \frac{4 x + 1}{3 - 2 x^{2} - x} - \frac{3 x^{2}}{1 - x^{3}} + \frac{3 π s i n (π x)}{2 + c o s (π x)}) . \frac{1}{f (x)}

f (0) = \frac{\sqrt[3]{6}}{27}

\frac{f^{'} (0)}{f (0)} = (\frac{- 1}{2} - \frac{1}{3}) . \frac{1}{3} = \frac{- 5}{18}

f^{'} (0) = - \frac{5}{18} . f (0) = - \frac{5 \sqrt[3]{6}}{18.27}

Commented by mnjuly1970 last updated on 17/May/21

g r a t e f u l \dots .

Commented by mindispower last updated on 17/May/21

p l e a s u r