Answer
$f(0)=1
,\\\\
f(15)=2
,\\\\
f(-82)=\text{does not exist}
,\\\\
f(80)=3$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Substitute the given function value in $
f(t)=\sqrt[4]{t+1}
.$
$\bf{\text{Solution Details:}}$
If $
t=0
,$ then
\begin{array}{l}\require{cancel}
f(t)=\sqrt[4]{t+1}
\\\\
f(0)=\sqrt[4]{0+1}
\\\\
f(0)=\sqrt[4]{1}
\\\\
f(0)=1
.\end{array}
If $
t=15
,$ then
\begin{array}{l}\require{cancel}
f(t)=\sqrt[4]{t+1}
\\\\
f(15)=\sqrt[4]{15+1}
\\\\
f(15)=\sqrt[4]{16}
\\\\
f(15)=2
.\end{array}
If $
t=-82
,$ then
\begin{array}{l}\require{cancel}
f(t)=\sqrt[4]{t+1}
\\\\
f(-82)=\sqrt[4]{-82+1}
\\\\
f(-82)=\sqrt[4]{-81}
\text{ (not a real number)}
.\end{array}
If $
t=80
,$ then
\begin{array}{l}\require{cancel}
f(t)=\sqrt[4]{t+1}
\\\\
f(80)=\sqrt[4]{80+1}
\\\\
f(80)=\sqrt[4]{81}
\\\\
f(80)=3
.\end{array}
Hence,
\begin{array}{l}\require{cancel}
f(0)=1
,\\\\
f(15)=2
,\\\\
f(-82)=\text{does not exist}
,\\\\
f(80)=3
.\end{array}
