calculate-f-x-1-4-1-cos-x-2-1-9-1-cos-x-2-x-2k-pi-k-Z-f-min-Adapted-From-Instagram-

Question Number 162721 by mnjuly1970 last updated on 31/Dec/21

c a l c u l a t e

f (x) = \frac{1}{4 (1 + c o s (\frac{x}{2}))} + \frac{1}{9 (1 - c o s (\frac{x}{2}))} (x \neq 2 k π, k \in Z)

f_{m i n} = ?

A d a p t e d F r o m I n s t a g r a m

Answered by Ar Brandon last updated on 31/Dec/21

f (x) = \frac{1}{4 (1 + \cos \frac{x}{2})} + \frac{1}{9 (1 - \cos \frac{x}{2})}

= \frac{1}{{8 \cos}^{2} \frac{x}{4}} + \frac{1}{{18 \sin}^{2} \frac{x}{4}} = \frac{8 + {10 \sin}^{2} \frac{x}{4}}{{36 \sin}^{2} \frac{x}{2}}

g (x) = {36 \sin}^{2} \frac{x}{2} \Rightarrow g^{'} (x) = 18 \sin x = 0 \Rightarrow x = π

f (x) \min \Rightarrow g (x) ma x \Rightarrow x = π

f_{m i n} = \frac{8 + {10 \sin}^{2} \frac{π}{4}}{{36 \sin}^{2} \frac{π}{2}} = \frac{8 + 5}{36} = \frac{13}{36}

Answered by mr W last updated on 31/Dec/21

f (x) = \frac{1}{4 \times 2 \cos^{2} \frac{x}{4}} + \frac{1}{9 \times 2 \sin^{2} \frac{x}{4}}

f (x) = \frac{1}{8} (\frac{1}{1 - \sin^{2} \frac{x}{4}} + \frac{4}{9 \sin^{2} \frac{x}{4}})

f (t) = \frac{1}{8} (\frac{1}{1 - t} + \frac{4}{9 t}) w i t h t = \sin^{2} \frac{x}{4} \in (0, 1)

f^{'} (t) = 0 \Rightarrow \frac{1}{{(1 - t)}^{2}} - \frac{4}{9 t^{2}} = 0

5 t^{2} + 8 t - 4 = 0

(5 t - 2) (t + 2) = 0

\Rightarrow t = \frac{2}{5}, - 2 (r e j e c t e d)

f_{m i n} = \frac{1}{8} (\frac{1}{1 - \frac{2}{5}} + \frac{4}{9 \times \frac{2}{5}}) = \frac{25}{72} ✓

Commented by mnjuly1970 last updated on 31/Dec/21

b r a v o s i r W n i c e s o l u t i o n

a s a l w a y s \dots g r a t e f u l \dots

Answered by mnjuly1970 last updated on 31/Dec/21

f (x) = \frac{1}{8} + \frac{1}{8} {t a n}^{2} (\frac{x}{4}) + \frac{1}{18} + \frac{1}{18} {c o t}^{2} (\frac{x}{4})

⩾ \frac{13}{72} + 2 \sqrt{\frac{1}{8} . \frac{1}{18}} = \frac{13}{72} + \frac{1}{6}

f_{m i n} = \frac{25}{72}