Show that the function f(x)=3 x +10, x 0 is decreasing. - Vidyadip

Question

Show that the function $f(x)=\frac{3}{x}+10, x \neq 0$ is decreasing.

✓

Answer

$
\begin{aligned}
& f(x)=\frac{3}{x}+10 \\
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{3}{x}+10\right)=3\left(-\frac{1}{x^2}\right)+0 \\
& \quad=-\frac{3}{x^2}
\end{aligned}
$
$
\begin{aligned}
& \because x \neq 0 \quad \therefore x^2>0 \quad \therefore \frac{3}{x^2}>0
\end{aligned}
$
$
\begin{aligned}
& \because x \neq 0 \quad \therefore x^2>0 \quad \therefore \frac{3}{x^2}>0 \\
& \therefore-\frac{3}{x^2}<0 \\
& \therefore f ^{\prime}( x )<0 \text { for all } x \in R , x \neq 0
\end{aligned}
$
Hence, the function $f$ is decreasing for all $x \in R$, where $x \neq 0$.

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