First slide

Maxima and minima

Question

If $y = x - \log (1 + x)$ , then minimum value of $y$ is

Moderate

A

1
B

0
C

-1
D

$\frac{1}{2}$

Solution

$y^{'} (x) = 1 - \frac{1}{1 + x} = \frac{x}{1 + x}$ The domain $y$ is $x > - 1 . so y^{'} (x) = 0 \Rightarrow x = 0$
$y^{''} (x) = \frac{1}{(1 + x)^{2}} > 0$ . Hence y has min at $x = 0$ and $y_{\min} =$ $0 - \log 1 = 0$

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