Local Extrema

Students were tasked with finding the local extrema of the function: $$f(t)=3t^4−4t^3,$$ where $t \in \mathbb{R}$.

All of them correctly found the first derivative: $$ f'(t)=12t^3−12t^2, $$ and set it equal to zero to find the critical points, obtaining the equation: $$ 12t^3−12t^2=0. $$ From this point, each student proceeded slightly differently.

Adam rewrote the equation as: $$ 12t^3=12t^2. $$ By dividing both sides by $t^2$, he obtained the equation $12t =12$, which has a single solution $t =1$. He then concluded that the only extremum of the function $f $ is at this point.

Bob factored the equation $12t^3−12t^2=0$ as: $$ 12t^2(t −1)=0 $$ and identified two solutions: $t_1=1$ and $t_2=0$. Bob then concluded that the function $f$ has local extrema at both these points.

David proceeded in a similar way to Bob, factoring the equation and obtaining two solutions, $t_1=1$ and $t_2=0$. To verify if these points are indeed extrema, he calculated the second derivative: $$ f''(t)=36t^2−24t $$ He substituted the critical points $t_1=1$ and $t_2=0$ into the second derivative: $$ f''(1)=36\cdot 1^2−24 \cdot 1=12>0 $$ which means that there is a local minimum at point $1$ and $$ f''(0)=36 \cdot 02−24 \cdot 0=0 $$

which means that there is no local extremum at point $0$.

Ema also found the two critical points $t_1=1$ and $t_2=0$ by factoring. She reasoned as follows:

Around $t_1=1$ the sign of the first derivative changes indicating that $f $ has a local extremum at point $1$. More precisely, on the interval $(0,1)$, the first derivative of $f$ is negative, and on the interval $(1,+\infty)$, it is positive. Because $f$ is continuous at point $1$, it holds that $f$ is decreasing on $[ 0,1 ]$ and it is increasing on $[ 1,+\infty)$. This means there is a local minimum at point $1$.

On the other hand, around $t_2=0$ the sign of the first derivative does not change, so $f$ does not have a local extremum at point $0$.

Which students did not make any mistakes?