Show that the function given by $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

Q: Show that the function given by $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

Explanation: Let $x_1$ and $x_2$ be any two real numbers such that $x_1 0$ for all $x \in \mathbb{R}$, the function is strictly increasing on $\mathbb{R}$.

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Accepted Answer

Explanation: Let $x_1$ and $x_2$ be any two real numbers such that $x_1 < x_2$.
Then,

$$f(x_1) = 3x_1 + 17 < 3x_2 + 17 = f(x_2)$$

$$\Rightarrow f(x_1) < f(x_2)$$

Hence, $f$ is strictly increasing on $\mathbb{R}$.

$ \bf Alternate Method:$

Differentiate the function:

$$f'(x) = \dfrac{d}{dx}(3x + 17) = 3$$

Since $f'(x) = 3 > 0$ for all $x \in \mathbb{R}$, the function is strictly increasing on $\mathbb{R}$.

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