Let f be a real valued function, defined on R - -1,1 and given by f(x)=3 log e|(x-1/x+1)|-(2/x-1) Then in which of the following intervals, function f ( x ) is increasing?

Question

Let f be a real valued function, defined on R - -1,1 and given by f(x)=3 log e|(x-1/x+1)|-(2/x-1) Then in which of the following intervals, function f ( x ) is increasing? (A) (-∞,-1) ∪([(1/2), ∞)- 1 ) (B) (-∞, ∞)- -1,1  (C) (-1, (1/2)] (D) (-∞, (1/2)]- -1

Tardigrade Admin · Accepted Answer

f ( x )=3 ln ( x -1)-3 ℓ n ( x +1)-(2/ x -1) f'( x )=(3/ x -1)-(3/ x +1)+(2/( x -1)2) f '( x )=(4(2 x -1)/( x -1)2( x +1)) f'( x ) ≥ 0 ⇒ x ∈(-∞,-1) ∪[(1/2), 1) ∪(1, ∞)

Q. Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by
$f (x) = 3 lo g_{e} ∣ ∣ \frac{x - 1}{x + 1} ∣ ∣ - \frac{2}{x - 1}$
Then in which of the following intervals, function $f (x)$ is increasing?

$(- \infty, - 1) \cup ([\frac{1}{2}, \infty) - {1})$

$(- \infty, \infty) - {- 1, 1}$

$(- 1, \frac{1}{2}]$

$(- \infty, \frac{1}{2}] - {- 1}$

$f (x) = 3 ln (x - 1) - 3 ℓ n (x + 1) - \frac{2}{x - 1}$
$f^{'} (x) = \frac{3}{x - 1} - \frac{3}{x + 1} + \frac{2}{( x - 1 ) ^{2}}$
$f^{'} (x) = \frac{4 ( 2 x - 1 )}{( x - 1 ) ^{2} ( x + 1 )}$
$f^{'} (x) \geq 0$
$\Rightarrow x \in (- \infty, - 1) \cup [\frac{1}{2}, 1) \cup (1, \infty)$

Q. Let f be a real valued function, defined on R−{−1,1} and given by f(x)=3loge​∣∣​x+1x−1​∣∣​−x−12​ Then in which of the following intervals, function f(x) is increasing?

(−∞,−1)∪([21​,∞)−{1})

(−∞,∞)−{−1,1}

(−1,21​]

(−∞,21​]−{−1}