A polynomial function $f(x)$ is such that $f'(4)=f''(4)=0$ and f(x) has minimum value 10 at x = 4. Then,

$f(x)=4+(x-4)^4$ $f(x)=10+(x-4)^4$ $f(x)=(x-4)^4$ none of these

$f(x)=10+(x-4)^4$

We have, $f^{\prime}(4)=0$ and $f^{\prime \prime}(4)=0$. ∴  $f(x)=(x-4)^n+\lambda$, where $n \geq 3$. It is given that $f(x)$ has minimum at $x=4$. Therefore, $n=4$. So, $f(x)=(x-4)^4+\lambda$ Now, $f(4)=10 \Rightarrow 10=\lambda$ Hence, $f(x)=(x-4)^4+10$