A differentiable function $f\left(x\right)$  has a relative minimum at x = 0. Then the function $y=f\left(x\right)+ax+b$  has a relative minimum at $x=0$  for

Since $f\left(x\right)$   has a relavtive minimum at $x=0$ , $f\text{'}\left(0\right)=0$   and   $f"\left(0\right)>0$.

if the function   $y=f\left(x\right)+ax+b$  has a relative minimum at $x=0$ , then

$$\begin{gathered} \frac{dy}{dx}=0 \text{at }x=0 \\ \Rightarrow \text{ f'}\left(x\right)+a=0 \text{for }x=0 \\ \Rightarrow f^{\mathrm{\prime }}\left(0\right)+a=0 \end{gathered}$$    

$0+a=0$  $$\begin{gathered} [\because f^{\mathrm{\prime }}(0)=0] \\ \Rightarrow a=0 \end{gathered}$$

 $$\begin{gathered} Now \frac{d^{2}y}{d{x}^2}=f^{\prime \prime }\left(x\right) \\ \Rightarrow { \left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}=f^{\prime \prime }\left(0\right)>0 \end{gathered}$$

$[\because f^{\prime \prime }(0)>0]$

Hence, $y$  has a relative minimum at $x=0$  if $a=0$  and $b$  can attain any real value.