Q. 13.7( 90 Votes )

# Show that the function given by f (x) = 3x + 17 is strictly increasing on R.

Answer :

Let x_{1} and x_{2} be any two numbers in R.

Then, we have,

x_{1} < x_{2}

⇒ 3x_{1} < 3x_{2}

⇒ 3x_{1} +17 < 3x_{2} +17

⇒ f(x_{1}) < f(x_{2})

Therefore, f is strictly increasing on R.

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