I came across a question on the applications of derivatives which asked me to calculate the maximum and the minimum values of the function $y=f(x)$ which is represented by the following equations:
$x=t^{5\ }-5t^3-20t+7$
$y=4t^3-3t^2-18t+3$
$|t|<2$
I obtained $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6}{5}\frac{(2t-3)(t+1)}{(t^2-4)(t^2+1)}$ which is equal to $0$ at $t=-1,$$\frac{3}{2}$
Since $\frac{\mathrm{d}y}{\mathrm{d}x} >0$ for $t\to-1^+$ and $< 0$ for values where $t\to-1^-$, I concluded that $-1$ should be the point of minima, which according to my book is actually the point of maxima. Similar is the case with $t=\frac{3}{2}$ where the values of $\frac{\mathrm{d}y}{\mathrm{d}x}$ just above and below $t=\frac{3}{2}$ suggest that it is the point of maxima whereas it actually is where $y$ attains the minimum value.
If I perform the second derivative test I get results which are consistent with the answers given in my book as $\frac{\mathrm{d^2}y}{\mathrm{d}x^2}$ at $t=-1 < 0$ which implies that it is indeed the point where $y$ is maximum and is equal to $14$.
Also $\frac{\mathrm{d^2}y}{\mathrm{d}x^2}$ at $t=\frac{3}{2} > 0$ which implies that it is the point where $y$ becomes minimum and is equal to $\frac{-69}{4}$.
Where am I going wrong? Is this happening because this function is parametric? I also plotted this curve on desmos, but it did not help me much.
Thanks for your suggestions/comments.
You have analyzed the behavior of $ \frac {dy}{dx}$ as $t \to -1$
What you really want is to analyze is $ \frac {dy}{dx}$ as $$x=t^{5 }-5t^3-20t+7$$ approaches to $31$.
See what happens to $ \frac {dy}{dx}$ as $x$ approaches to $31$ from right or left.