Question

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

• A. All $a$ and All $b$
• B. All $b,$ if $a=0$
• C. All $b>0$
• D. All $a>0$

Answer 1

The correct option is B All $b,$ if $a=0$

Since, $f\left(x\right)$ has a relative minimum at $x=0$. Therefore, $f^{\prime }\left(0\right)=0$ and $f^{\prime \prime }\left(0\right)>0$.
If the function $y=f\left(x\right)+ax+b$ has a relative minimum at $x=0$, then
$\frac{dy}{dx}=0$ at $x=0$
$\Rightarrow f^{\prime }\left(x\right)+a=0$ for $x=0$
$\Rightarrow f^{\prime }\left(0\right)+a=0$
$\Rightarrow 0+a=0$
$\Rightarrow a=0$ $[\because f^{\prime }(0)=0]$
Now, $\frac{d^{2}y}{d{x}^2}=f^{\prime \prime }\left(x\right)$
${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}=f^{\prime \prime }\left(0\right)>0$
$[\because f^{\prime \prime }(0)>0]$
Hence, $y$ has relative minimum at $x=0$, if $a=0$ and $b$ can attain any real value.