Let f be a real valued function, defined on R - -1, 1 and given by f(x) = 3 loge [| x - 1 …

Question

Let f be a real valued function, defined on R - {-1, 1} and given by

f(x) = 3 loge $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$.

Then in which of the following intervals, function f(x) is increasing?

Accepted Answer

f(x) = 3 loge $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$

$$f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0$$

$$ = {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0$$

$$ = {{2(3(x - 1) + (x + 1))} \over {{{(x - 1)}^2}(x + 1)}} = {{4(2x - 1)} \over {{{(x - 1)}^2}(x + 1)}} > 0$$

$$ \Rightarrow x \in ( - \infty , - 1) \cup \left[ {{1 \over 2},\infty ) - \{ 1\} } \right.$$

Solutions · 1