Question

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

Answer 1

Hint: We know that function is increasing when its differentiation is greater than zero that means its slope is positive. So here we have to differentiate the given function.

Complete step-by-step answer:
$f\left( x \right) = 3x + 17$ is strictly increasing on R
As we know that differentiation is also known as the slope of function
Then, on differentiating this function w.r.t x
$f’(x)$ = $\dfrac{d}{{dx}}\left( {3x + 17} \right) = 3$
$\therefore f’(x) = 3$
Here we can see that 3 > 0, so $f’(x) > 0$, for all $x \in R$
Since differentiation is greater than zero which means slope is greater than zero, thus the given function is strictly increasing

NOTE: Another easy way to find increasing or decreasing function by sketching the graph of function. A function is called monotonically increasing (also increasing or non-decreasing) if for all x and y such that $x \leqslant y$ one has $f(x) \leqslant f(y)$