Question
Find the values of $x$, such that $f(x)$ is increasing function:
$f(x)=x^2+2 x-5$

Answer

$
\begin{aligned}
& f(x)=x^2+2 x-5 \\
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(x^2+2 x-5\right) \\
& =2 x+2 \times 1-0 \\
& =2 x+2
\end{aligned}
$
$f$ is increasing, if $f^{\prime}(x)>0$
i.e. if $2 x+2>0$
i.e. if $2 x>-2$
i.e. if $x>-1$, i.e. $x \in(-1, \infty)$
$\therefore f$ is increasing, if $x >-1$, i.e. $x \in(-1, \infty)$

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