Question

If $f(x)=\begin{cases}\displaystyle\lim _{t \rightarrow \infty}\left(6^{t}+10^{t}\right)^{\frac{1}{t}}, & x=0 \\ x^{2}+4 x+ C & , x \neq 0\end{cases}$ is a continuous function, then minimum value of $f(x)$ is

Answer 1

$f(0)=10 \displaystyle\lim _{t \rightarrow \infty}\left(\left(\frac{6}{10}\right)^{t}+1\right)^{\frac{1}{t}}=10$
$\displaystyle\lim _{x \rightarrow 0} f(x)=x^{2}+4 x+ c=c$
$c=10$
$f(x)=x^{2}+4 x+10=(x+2)^{2}+6$
minimum value $=6$