Minimum-value-of-function-f-x-16x-2-cos-2-x-4-x-cos-x-where-pi-lt-x-lt-0-

Question Number 115332 by bobhans last updated on 25/Sep/20

M i n i m u m v a l u e o f f u n c t i o n

f (x) = \frac{16 x^{2} \cos^{2} x + 4}{x \cos x} w h e r e - π < x < 0

Commented by bemath last updated on 25/Sep/20

\Leftrightarrow f (x) = 16 x \cos x + 4 x^{- 1} \sec x

f^{'} (x) = 16 \cos x - 16 x \sin x + (- 4 x^{- 2} \sec x + 4 x^{- 1} \sec x \tan x) = 0

4 \cos x - 4 x \sin x = x^{- 1} \sec x (x^{- 1} - \tan x)

4 \cos x - 4 x \sin x = \frac{1}{x \cos x} (\frac{1}{x} - \frac{\sin x}{\cos x})

4 \cos x - 4 x \sin x = \frac{\cos x - x \sin x}{x \cos x}

(\cos x - x \sin x) (4 - \frac{1}{x \cos x}) = 0

(\cos x - x \sin x) (\frac{4 x \cos x - 1}{x \cos x}) = 0

{\begin{cases} \cos x = x \sin x \Rightarrow \tan x = \frac{1}{x} \\ 4 x \cos x = 1 \Rightarrow \cos x = \frac{1}{4 x} \end{cases}

(1) f o r \tan x = \frac{1}{x} \Rightarrow \tan^{2} x = \frac{1}{x^{2}}

\sec^{2} x = 1 + \frac{1}{x^{2}} = \frac{x^{2} + 1}{x^{2}}

\cos^{2} x = \frac{x^{2}}{x^{2} + 1} \Rightarrow \cos x = \pm \sqrt{\frac{x^{2}}{x^{2} + 1}}

f (x) = \frac{16 x^{2} (\frac{x^{2}}{x^{2} + 1}) + 4}{\pm x \sqrt{\frac{x^{2}}{x^{2} + 1}}} = \frac{16 x^{4} + 4 x^{2} + 4}{\pm x^{2} \sqrt{x^{2} + 1}}

(2) f o r \cos x = \frac{1}{4 x}

f (x) = \frac{16 x^{2} (\frac{1}{16 x^{2}}) + 4}{x (\frac{1}{4 x})} = \frac{5}{(\frac{1}{4})} = 20

Commented by soumyasaha last updated on 25/Sep/20

f (x) = 16 xcosx + \frac{4}{xcosx}

Now, - π < x < 0 \Rightarrow cosx < 0 and x \neq \frac{π}{2}

\Rightarrow xcosx > 0

We know, A . M . ⩾ G . M .

\Rightarrow \frac{16 cosx + \frac{4}{xcosx}}{2} ⩾ \sqrt{16 cosx . \frac{4}{xcosx}}

\Rightarrow \frac{f (x)}{2} ⩾ \sqrt{64}

\Rightarrow f (x) ⩾ 16

∴ minimum value of f (x) is 16 .

Commented by bemath last updated on 25/Sep/20

Answered by soumyasaha last updated on 25/Sep/20

Commented by bemath last updated on 25/Sep/20

w h e n x = ? s i r f o r f (x) m i n i m u m

i {d o n}^{'} t h a v e e x a c t v a l u e

Answered by MJS_new last updated on 25/Sep/20

\cos - \frac{π}{2} = 0 \Rightarrow absolute minima / maxima

{don}^{'} t exist

you can only find local minima / maxima

Answered by TANMAY PANACEA last updated on 25/Sep/20

\frac{16 t^{2} + 4}{t} t = x c o s x

4 (4 t + \frac{1}{t})

4 [{(2 \sqrt{t} - \frac{1}{\sqrt{t}})}^{2} + 2 \times 2 \sqrt{t} \times \frac{1}{\sqrt{t}}]

4 [{(2 \sqrt{t} - \frac{1}{\sqrt{t}})}^{2} + 4]

16 + 4 {(2 \sqrt{t} - \frac{1}{\sqrt{t}})}^{2} \to m i n v a l u e = 16

Answered by MJS_new last updated on 25/Sep/20

local maximum at (\begin{matrix} - π \\ 16 π + \frac{4}{π} \end{matrix})

local minimum at (\begin{matrix} \approx - 1.84520 \\ 16 \end{matrix})

local maximum at (\begin{matrix} \approx - 1.09801 \\ - 16 \end{matrix})

local minimum at (\begin{matrix} \approx - . 860334 \\ \approx - 16.1064 \end{matrix})

local maximum at (\begin{matrix} \approx - . 610031 \\ - 16 \end{matrix})

absolute \min / \max {doesn}^{'} t exist because

- \infty < f (x) < \infty

proof :

\lim_{x \to {(- \frac{π}{2})}^{-}} f (x) = + \infty

\lim_{x \to {(- \frac{π}{2})}^{+}} f (x) = - \infty

\lim_{x \to 0^{-}} f (x) = - \infty

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