Answer
The curve is concave up for the $t$-intervals: $t < - \frac{1}{{{2^{1/3}}}}$ and $t>0$.
Work Step by Step
We have
$x\left( t \right) = {t^2}$ and $y\left( t \right) = {t^4} - 4t$.
So,
$x'\left( t \right) = 2t$, $x{\rm{''}}\left( t \right) = 2$,
$y'\left( t \right) = 4{t^3} - 4$, $y{\rm{''}}\left( t \right) = 12{t^2}$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = \frac{{1 + 2{t^3}}}{{{t^3}}}$.
By Theorem 1 of Section 4.4, the curve is concave up if $\frac{{{d^2}y}}{{d{x^2}}} > 0$.
So we solve $\frac{{1 + 2{t^3}}}{{{t^3}}} > 0$.
Case 1: $t<0$, then $1+2 t^3<0$ such that $\frac{{1 + 2{t^3}}}{{{t^3}}} > 0$.
$1 + 2{t^3} < 0$,
$2{t^3} < - 1$, ${t^3} < - \frac{1}{2}$.
Thus, the solution is $t < - \frac{1}{{{2^{1/3}}}}$.
Case 2: $t>0$, then $1+2 t^3>0$ such that $\frac{{1 + 2{t^3}}}{{{t^3}}} > 0$.
$1 + 2{t^3} > 0$,
$2{t^3} > - 1$, ${t^3} > - \frac{1}{2}$.
$t > - \frac{1}{{{2^{1/3}}}}$.
Thus, the solution is $t>0$.
From case 1 and case 2 we conclude that the curve is concave up for the $t$-intervals: $t < - \frac{1}{{{2^{1/3}}}}$ and $t>0$.
