Question
Without using the derivative show that the function f(x) = 7x - 3 is strictly increasing function on R.
✓
Answer
Here,
$\text{f}(\text{x})=7\text{x}-3$
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $\text{x}_1<\text{x}_2.$ Then,
$\text{x}_1<\text{x}_2$
$\Rightarrow7\text{x}_1<7\text{x}_2$ $[\because\ 7>0]$
$\Rightarrow7\text{x}_1-3<7\text{x}_2-3$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$
$\therefore\ \text{x}_1<\text{x}_2\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in\text{R}$
So, f(x) is strictly increasing on R.
$\text{f}(\text{x})=7\text{x}-3$
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $\text{x}_1<\text{x}_2.$ Then,
$\text{x}_1<\text{x}_2$
$\Rightarrow7\text{x}_1<7\text{x}_2$ $[\because\ 7>0]$
$\Rightarrow7\text{x}_1-3<7\text{x}_2-3$
$\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2)$
$\therefore\ \text{x}_1<\text{x}_2\Rightarrow\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in\text{R}$
So, f(x) is strictly increasing on R.
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