# Show that the fun

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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