Find-the-maximum-value-of-f-x-3-2cosh-ln-x-3-

Question Number 100370 by Rio Michael last updated on 26/Jun/20

Find the maximum value of f (x) = \frac{3}{2 \cosh (\ln x) + 3}

Commented by bemath last updated on 26/Jun/20

y_{\max} = 0.6

Commented by mr W last updated on 26/Jun/20

m a x i m u m f (x) m e a n s m i n i n u m \cosh (\ln x),

w h i c h i s 1 w h e n \ln x = 0 o r x = 1 .

m a x . f (x) = f (1) = \frac{3}{2 \times 1 + 3} = \frac{3}{5}

Answered by MJS last updated on 26/Jun/20

\cosh \ln x = \frac{1}{2} (x + \frac{1}{x})

f (x) = \frac{3 x}{x^{2} + 3 x + 1}

f^{'} (x) = - \frac{3 (x^{2} - 1)}{{(x^{2} + 3 x + 1)}^{2}}

f^{'} (x) = 0 \Rightarrow x = \pm 1

f^{″} (x) = \frac{6 (x^{3} - 3 x - 3)}{{(x^{2} + 3 x + 1)}^{3}}

f^{″} (- 1) = 6 > 0 \Rightarrow \min at x = - 1; y = 3

f^{″} (1) = - \frac{6}{25} < 0 \Rightarrow \max at x = 1; y = \frac{3}{5}

but these are only local

- \infty < f (x) < + \infty

f (x) is not defined for x = - \frac{3}{2} \pm \frac{\sqrt{5}}{2}

Commented by bemath last updated on 26/Jun/20

sir \cosh (\ln x) = \frac{1}{2} (x + \frac{1}{x}) i t i s f r o m s e r i e s ?

Commented by MJS last updated on 26/Jun/20

no; from definition

\cosh x = \frac{e^{x} + e^{- x}}{2}

\sinh x = \frac{e^{x} - e^{- x}}{2}

\cos x = \frac{e^{i x} + e^{i x}}{2}

\sin x = \frac{e^{i x} - e^{- i x}}{2 i}

Commented by bemath last updated on 26/Jun/20

oo thank you sir

Commented by Rio Michael last updated on 26/Jun/20

thank {you}^{'} all

Commented by mr W last updated on 26/Jun/20

M J S s i r :

i t h i n k \cosh (\ln x) a n d \frac{1}{2} (x + \frac{1}{x}) a r e

n o t e x a c t l y t h e s a m e .

t h e d o m a i n o f \cosh (\ln x) i s x > 0,

b u t t h e d o m a i n o f \frac{1}{2} (x + \frac{1}{x}) i s x < 0

a n d x > 0 .

t h e r e f o r e \frac{3}{2 \cosh (\ln x) + 3} h a s r e a l l y

a g l o b a l m a x i m u m \frac{3}{6} .

Commented by MJS last updated on 26/Jun/20

I think \cosh \ln x is defined for x \in R ∖ {0}

because \ln (- ∣ x ∣) = π i + \ln ∣ x ∣

and e^{π i + \ln ∣ x ∣} = - ∣ x ∣ and e^{- (i π + \ln x)} = - \frac{1}{∣ x ∣}

\Rightarrow \cosh \ln (- ∣ x ∣) = - \frac{x^{2} + 1}{2 ∣ x ∣}

Facebook Tweet Pin