Q2 of 30 Page 6

Show that the function given by f(x) = 7x – 3 is strictly increasing on R.

f(x) = 7x – 3


Now f(x) is strictly increasing iff f’(x) > 0


Let us calculate f’(x)


Differentiate f(x) with respect to x


⇒ f’(x) = 7


Hence f’(x) is constant


7 > 0 for all values of x hence f’(x) > 0


As f’(x) > 0, f(x) is strictly increasing


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