Question

$f\left(c\right)$ is minimum value of $f\left(x\right)$, if-

• A. $f^{\prime }\left(c\right)=0$, $f^{\prime \prime }\left(c\right)>0$
• B. $f^{\prime }\left(c\right)=0$, $f^{\prime \prime }\left(c\right)<0$
• C. $f^{\prime }\left(c\right)\ne 0$, $f^{\prime \prime }\left(c\right)=0$
• D. $f^{\prime }\left(c\right)<0$, $f^{\prime \prime }\left(c\right)>0$

Answer 1

The correct option is A $f^{\prime }\left(c\right)=0$, $f^{\prime \prime }\left(c\right)>0$

Consider a function $y=f\left(x\right)$
Now for the function to have a minimum value at $c$, the slope of that function at $c$ should be $0.$
Or
slope
$=f^{\prime }(x{)}_{x=c}$
$=0$
Or
$f^{\prime }\left(c\right)=0$ ...(i)
Now applying the second derivative test which says,
$f^{\prime \prime }\left(\alpha \right)>0$ minimum at $\alpha$
$f^{\prime \prime }\left(\alpha \right)<0$ maximum at $\alpha$.
Hence $f"\left(c\right)>0$.