# 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 \\

\ $

Last updated date: 23rd Mar 2023

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Answer

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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$f'\left( x \right) = a$.

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$f'\left( x \right) = a$.

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