Question

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

Answer 1

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 $f^{\prime }(x)>0$.

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

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

We know that for any increasing function $f^{\prime }(x)$ should be greater than zero i.e. $f^{\prime }(x)>0$.
From the given function we can say that $f^{\prime }\left(x\right)=a$
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$f^{\prime }\left(x\right)=a$.