Question
Show that function $f(x)=7 x^2-3, x>0$ is increasing function.

Answer

$
f(x)=7 x^3-3, x>0
$
then$
f^{\prime}(x)=14 x-0
$$\forall x>0$
$\Rightarrow \quad f^{\prime}(x)>0, \quad \forall x>0$
hence function $f(x)$ is increasing function because $f^{\prime}(x)>0$.

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