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The given following function $f(x) = ax + b$ is strictly increasing for all real $x$ if
$\
  A) a > 0 \\
  B) a < 0 \\
  C) a = 0 \\
  D) a \leqslant 0 \\
\ $

Answer Verified Verified
Hint: Suppose that a function y=f(x) is a differentiable function. In order for the function to be strictly increasing it is necessary and sufficient that it has to follow a condition where $f'(x) > 0$.

Here the given function is $f(x) = ax + b$.

To find the condition for given increasing function we have to find $f'(x)$
So, if we differentiate the given function $f(x)$ with respective $x$ we get
 $f(x) = ax + b$
$f'\left( x \right) = a$

We know that for any increasing function $f'(x)$ should be greater than zero i.e. $f'(x) > 0$.
From the given function we can say that $f'\left( x \right) = a$
So from this we can say that for the given function $f(x) = ax + b$ is strictly increasing function for all real x when$a > 0$, where$f'\left( x \right) = a$.