f-x-4x-2-1-2x-2-1-prove-that-1-f-x-2-

Question Number 130354 by greg_ed last updated on 24/Jan/21

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

prove that 1 ⩽ f (x) ⩽ 2

Answered by MJS_new last updated on 24/Jan/21

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

minimum of \frac{1}{2 x^{2} + 1} is 0 (with x = \pm \infty)

maximum of \frac{1}{2 x^{2} + 1} is 1 (with x = 0)

\Rightarrow

minimum of f (x) is 2 - 1 = 1

maximum if f (x) is 2 - 0 = 2

Answered by ajfour last updated on 24/Jan/21

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

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

\Rightarrow 1 ⩽ f (x) ⩽ 2

Answered by nueron last updated on 24/Jan/21

Answered by Olaf last updated on 24/Jan/21

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

\Rightarrow f (x) ⩽ 2 (trivial)

f (x) - 1 = \frac{4 x^{2} + 1}{2 x^{2} + 1} - 1 = \frac{2 x^{2}}{2 x^{2} + 1} ⩾ 0

\Rightarrow f (x) ⩾ 1 (trivial)

Answered by john_santu last updated on 25/Jan/21

l e t f (x) = y

\Rightarrow 2 x^{2} y - 4 x^{2} + y - 1 = 0

\Rightarrow (2 y - 4) x^{2} + y - 1 = 0

D = 0 - 4 . (y - 1) (2 y - 4) ⩾ 0

\Rightarrow (y - 1) (2 y - 4) ⩽ 0

\Rightarrow 1 ⩽ y ⩽ 2 \Rightarrow 1 ⩽ f (x) ⩽ 2