Find the intervals in which the function $f(x) = \log (1 + x) −\frac{x}{1+x}$ is strictly increasing or strictly decreasing.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Strictly decreasing on $(−1,0]$ and strictly increasing on $[0,∞)$.
Given $f(x) = \log (1 + x) -\frac{x}{1+x}$. ...(i)
For $D_f, 1+x>0$ and $1+x≠0⇒ x > -1$ and $x≠-1⇒x>-1$
⇒ domain of the function = (-1, ∞).
Differentiating the given function w.r.t. x, we get
$f'(x)=\frac{1}{1+x}.1 − \frac{(1 + x).1 − x (0 + 1)}{(1 + x)^2}-\frac{1}{1+x}-\frac{1}{(1+x)^2}=\frac{1+x-1}{(1+x)^2}=\frac{x}{(1+x)^2}$
Now $f'(x) > 0$ iff $\frac{x}{(1+x)^2}> 0⇒ x > 0, x ≠ -1$ $(∵(1 + x)^2>0,∀x∈R,x ≠ -1)$
= f(x) is strictly increasing in $[0, ∞)$.
And $f'(x) < 0$ iff $\frac{x}{(1+x)^2}<0⇒x<0,x≠-1$ but $x ∈(-1, ∞)$
= f(x) is strictly decreasing in $(-1, 0]$.
Hence, the given function is strictly increasing in [0, ∞) and strictly decreasing in (-1, 0].