Answer
$t= 3,-1$
Work Step by Step
Given $$f(t)=t-4 \sqrt{t+1}$$
Since
\begin{align*}
f'(t)&= 1-4\frac{1}{2\sqrt{t+1}}\\
&=1-\frac{2}{\sqrt{t+1}}\\
&=\frac{\sqrt{t+1}-2}{\sqrt{t+1}}
\end{align*}
Then to find the critical points
\begin{align*}
f'(t)&=0\\
\frac{\sqrt{t+1}-2}{\sqrt{t+1}}&=0\\
\sqrt{t+1}-2&=0\\
t+1&=4
\end{align*}
Hence $f(t)$ has a critical point at $t=3$
In addition, $t=-1$ is also a critical point because $f'(t)$ is undefined there (division by zero).
