The function $y=f(x)$ is represented parametrically by $x=t^5-5t^3-20t+7$ and $y=4t^3-3t^2-18t+3$ $(-2\lt t\lt2)$. The minimum of $y=f(x)$ occurs at Question:
The function $y=f(x)$ is represented parametrically by $x=t^5-5t^3-20t+7$ and $y=4t^3-3t^2-18t+3$ $(-2\lt t\lt2)$. The minimum of $y=f(x)$ occurs at $t=...$
My Attempt:
For extrema, I am calculating $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{12t^2-6t-18}{5t^4-15t^2-20}=\frac{6(2t^2-t-3)}{5(t^4-3t^2-4)}=\frac{6(2t-3)(t+1)}{5(t^2-4)(t^2+1)}$
Around $t=\frac32, f'(x)$ sign changes from positive to negative. So $t=\frac32$ is local maxima.
Around $t=-1, f'(x)$ sign changes from negative to positive. So $t=-1$ is local minima.
But the answer given is $\frac32$
What's wrong in my approach?
 A: I have plotted the parametrised function for $-2\leq t\leq 2$ and i got 
with $A=f\left(\frac{3}{2}\right)$, thus there is indeed a minimum. The question is why and the point is that the parametrization moves in one specific direction, in this case I took $B=f(1)$ and $C=f(1.25)$ and then you see that you move to the left along the graph of $f$ if $t$ goes from $1$ to $\frac{3}{2}$.

Thus for smaller $t$ (close to $\frac{3}{2}$) is $f'$ positive (but that is in $x$ value on the right hand side of the point $A$ in the xy-plane) and thus for larger values of $t$ (close to $\frac{3}{2}$) are you on the left hand side of the point $A$ in the xy-plane and is $f'$ negative and thus is it a local minima.
A: 
The function $y=f(x)$ is represented parametrically by $x=t^5-5t^3-20t+7$ and $y=4t^3-3t^2-18t+3$ $(-2\lt t\lt2)$. The minimum of $y=f(x)$ occurs at $t=...$

Alternative approach:
Since you are (in effect) given $~y = g(t) ~: ~-2 < t < 2,$ 
and since you are asked for the value of $(t)$ [rather than the value of $(x)$] where $(y)$ is minimized, you can totally ignore the variable $(x)$.
Further, for my approach, it is necessary to take the first derivative of $~\dfrac{dy}{dt}.~$  However, beyond that, I have a choice:

*

*After identifying the value(s) of $(t)$ where $~\dfrac{dy}{dt} = 0,~$ I can check whether the 2nd derivative is positive or negative.


*Or, I can ignore the 2nd derivative, and simply compute the value of $(y)$ at each of the points where the first derivative is $(0)$, as well as the boundary points $(t)$ approaching $\pm 2.$
Which choice I make, the 1st method above, or the 2nd method above will depend on

*

*How easy it is to interrogate the 2nd derivative.


*Whether there is more than one value for $(t)$ such that $\dfrac{dy}{dt} = 0.$
$y = g(t) = 4t^3-3t^2-18t+3 \implies g'(t) = 12t^2 - 6t - 18.$
So,
$~\displaystyle g'(t) = 0 \implies (2t^2 - t - 3) = 0 \implies 
t = \frac{1}{4} \left[1 \pm \sqrt{1 + 24}\right] \implies $
$\displaystyle t \in \left\{ ~\frac{3}{2}, -1 ~\right\}.$
Although the 2nd derivative, $g''(t)$ would be easy to interrogate, I will take the simpler choice of evaluating $g(t)$ at each critical point, as well as when $(t)$ approaches the boundary points.  I make this choice because there is more than one critical point.  Therefore, an interrogation based solely on the 2nd derivative would not be conclusive anyway.
Relying heavily on the fact that $g(t)$ is a continuous function, I will interrogate the value of $g(t)$ as $(t)$ approaches each of the boundary points, by explicitly computing $g(t)$ at these boundary points.

*

*$t = \dfrac{3}{2} \implies g(t) = -\dfrac{69}{4}.$


*$t = -1 \implies g(t) = 14.$


*$t = 2 \implies g(t) = -13.$


*$t = -2 \implies g(t) = -5.$
Therefore, $y = g(t)$ is minimized at $t = \dfrac{3}{2}$, 
and this minimum is $y = -\dfrac{69}{4}.$
A: When I plot the parametric equation in desmos, I see a local min at approximately $x = -32$.  Then, I solve the equation $t^5-5t^3-20t+7 = -32$.  One of the roots is very close to $3/2$.  So, I think $3/2$ is definitely correct. There is more to this problem than meets the eye!
A: A short answer is that when we apply parametric first derivatives to a curve, we also have to give attention to the direction "traced" along the curve by increasing $ \ t \ \ $ (the same is true for integration of "area under the curve" -- I learned this the hard way...).  We don't generally give this a thought for single-variable functions with $ \ f'(x) \ $ because we (usually) take $ \ x \ $ as increasing "to the right".  However, as illustrated by René Bruin, increasing $ \ t \ $ traces this portion of the curve "right to left", so having the first derivative increase from negative to positive for $ \ t \ $ increasing while "passing through" $ \ -1 \ $ is equivalent to the derivative decreasing from positive to negative as $ \ \mathbf{x} \ $ decreases as it passes through $ \ 32 \ \ . $  Similarly, $ \ x \ $ is still being traced "right to left" near $ \ -\frac{1033}{32} \ $ as $ \ t \ $ increases through $ \ \frac32 \ \ ; \ $ so the first derivative is "actually" changing from negative to positive there, making it a "minimum" turning-point on the curve.  (This has been also been said in different ways by the other posters.)
With some computational aid, we can confirm this by finding the parametric second derivative (which is the actual point of my posting):
$$ \frac{d^2 y}{dx^2} \ \ = \ \ \frac{\frac{d}{dt} \ [ \ \frac{dy}{dx} \ ]}{\frac{dx}{dt}} \ \ = \ \  \large{\frac{\frac{d}{dt} \ [ \ \frac{6 \ · \ (2t^2 \ - \ t \ - \ 3)}{5 \ · \ (t^4 \ - \ 3t^2 \ - \ 4)} \ ]}{5·(t^4  -  3t^2  -  4)}}$$ $$ =  \ \ \frac{ \ {6·( \ 4t^5 \ - \ 3t^4 \ - \ 12t^3 \ + \ 3t^2 \ + \ 34t \ - 4 \ )}}{25·( \ t^4-3t^2-4 \ )^3} \ \ . $$
[It's understandable that you didn't want to work  this out "by hand".]
At $ \ t \ = \ -1 \ \ , \ $ we obtain $$ \frac{d^2 y}{dx^2} \ \ = \ \ \frac{ \ {6·( \ -30 \ )}}{25·( \ -6 \ )^3} \ \ = \ \ -\frac{1}{30} \ \ ; $$
since this is negative, the curve near this point is "concave downward" as we view it on the graph.  At the other turning-point $ \ ( t \ = \ \frac32 ) \ \ , $
$$ \frac{d^2 y}{dx^2} \ \ = \ \ \frac{ \ \frac{1365}{8}}{\frac{207025}{256}} \ \ = \ \ \frac{96}{455} \ \ , $$
so the curve is seen as "concave upward" in this neighborhood.
This comes out "right way 'round" because the parametric second derivative $ \  \large{\frac{\frac{d}{dt} \ [ \ \frac{dy}{dx} \ ]}{\frac{dx}{dt}}} \ $ "measures" the rate at which the first derivative changes relative to the rate at which $ \ x \ $ is changing as $ \ t \ $ increases.
