Question

The function $f$ is such that $f^{\prime }\left(4\right)=f^{\prime \prime }\left(4\right)=0$ and $f$ has minimum value $10$ at $x=4$. Then $f\left(x\right)=k+{(x-m)}^n$, where $k+m+n=$

Answer 1

Since $f^{\prime }\left(4\right)=f^{\prime \prime }\left(4\right)=0,$ therefore, $f\left(x\right)={(x-4)}^n+k,$ where $n\ge 3$

But $f$ has minimum at $x=4,$ so $n=4.$

$\therefore f\left(x\right)={(x-4)}^4+k.$

Since $f\left(4\right)=10,$ therefore, $k=10$

Thus, $f\left(x\right)={(x-4)}^4+10$

$\therefore k+m+n=10+4+4=18$