Prove that f(x) = ax + b, where a, b are constants and a < 0 is a… - Vidyadip

Question

Prove that $f(x) = ax + b,$ where $a, b$ are constants and $a < 0$ is an decreasing function on $R.$

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Answer

Here, $f(x) = ax + b$ Let $\text{x}_1,\text{x}_2\in\text{R}$
such that $x_1 < x_2.$
Then, $x_1 < x_2$
$\Rightarrow ax_1 > ax_2(\because\ \text{a}<0)$
$\Rightarrow ax_1 + b > ax_2 + b $
$\Rightarrow f(x_1) > f(x_2)$
Thus, $x_1 < x_2$
$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in\text{R}$
So, $f(x)$ is decreasing on $R.$

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