To find the critical points of $f$, we must examine its derivative: \begin{align*} f'(x) &= \frac{d}{dx} \left[2x^3 -3x^2 -12x +1 \right] \\ \\ &= (2)(3)x^2 -3(2)x -12 \\ \\ &= 6x^2 -6x -12 \end{align*} The critical points occur when $f'(c) = 0$. (Critical points could also occur where $f'(c)$ is undefined, but this polynomial is defined everywhere so we only have to consider when $f'(c) = 0$.)

\begin{align*} f'(c) = 6c^2 -6c -12 &= 0 \\ 6\left(x^2 – c -2 \right) &= 0\\ 6 (c -2)(c+1) &= 0\\ c &= 2, \, -1 \quad \cmark \end{align*}

Most people using the First Derivative Test find it helpful to create a small diagram like the one below to organize their thinking and results. We place the values of $x$ underneath the number line, marking the critical numbers we found earlier. Above the line we show information about $f'(x)$, starting with the fact that at these critical points $f’$ is zero.



The figure nicely illustrates that we need to consider the sign of $f’$ in three regions: (I) to the left of -1; (II) between -1 and 2; and (III) to the right of 2.

(I) An easy value to try to the left of -1 is $x =-2$:

\begin{align*} f'(-2) &= 6(-4)(-1) \\[8px] &> 0 \end{align*} We actually don’t care what the value of $f'(-2)$ is; we only care whether it’s positive or negative. Since it’s positive, we put a “+” above that region on our line:



We thus know that the function has a positive derivative, and so is increasing, on the interval $(-\infty, -1)$.

(II) A convenient value between -1 and 2 is zero:

\begin{align*} f'(0) &= 6(-2)(1) \\[8px] &< 0 \end{align*} Again, we don't care about the value specifically; without any further work we see that it's negative, and so we put a "–" above that region.



We thus know that the function has a negative derivative, and so is decreasing, on the interval $(-1, 2)$.

Furthermore, we now know that the critical point $x = -1$ is a relative maximum, according to the First Derivative Test, because $f’$ switches from positive to negative there.



(III) A convenient value to try to the right of 2 is $x=3$:

\begin{align*} f'(3) &= 6(1)(4) \\[8px] &>0 \end{align*} This value is positive, and so we put a “+” over that region on the line.



We now know that, after $x = 2$, the function always increases: it is increasing on $(2, \infty)$.

Furthermore, we conclude from the First Derivative Test that $x = 2$ is a relative minimum, since $f’$ switches from negative to positive there.



Summarizing our results, we conclude our First Derivative Tests:
$\bullet$ Since $f’$ switches from positive to negative at $x = -1$, this point is a local maximum. $\quad \cmark$
$\bullet$ Since $f’$ switches from negative to positive at $x = 2$, this point is a local minimum. $\quad \cmark$

For comparison, the figure below shows the graph of the function.

To use the Second Derivative Test we must compute the function’s second derivative. We start with the first derivative that we found in part (a):

\begin{align*} f'(x) &= 6x^2 -6x -12 \\[12px] f”(x) &= \dfrac{d}{dx} \left[6x^2 -6x -12 \right] \\[8px] &=12x – 6 \end{align*}

For the Second Derivative Test, we must determine whether $f”(c)$ is positive or negative (or zero) for each of our critical points. Recall from part (a) that our critical points are $c = -1, \, 2$:

(I) Let’s look first at $c = -1$, and determine whether $f”(-1)$ is positive or negative:

\begin{align*} f”(-1) &= 12(-1) – 6 \\[8px] &= -18 < 0 \end{align*} That is, the curve is concave down at $x = -1$, and so this critical point is a local maximum.



(II) Next let’s look first at $c = 2$, and determine whether $f”(2)$ is positive or negative:

\begin{align*} f”(2) &= 12(2) – 6 \\[8px] &= 18 > 0 \end{align*} That is, the curve is concave up at $x = 2$, and so this critical point is a local minimum.



The Second Derivative Test is thus a different way or arriving at the conclusions we reached in part (b):
$\bullet$ Since $f”$ is negative $x = -1$, this point is a local maximum. $\quad \cmark$
$\bullet$ Since $f”$ is positive at $x = 2$, this point is a local minimum. $\quad \cmark$

To find the critical points of $f$, we must examine its derivative: \begin{align*} f'(x) &= \frac{d}{dx} \left[e^{-x} \sin x \right] \\ \\ &= \left(\frac{d}{dx} e^{-x} \right) \sin x + e^{-x} \left(\frac{d}{dx} \sin x \right) \\ \\ &= \left(-e^{-x}\right) \sin x + e^{-x} (\cos x) \\ \\ &= \left(-e^{-x}\right) (\sin x – \cos x) \end{align*}

The critical points occur when $f'(c) = 0$. (We could also have critical points where $f'(c)$ is undefined, but $f'(x)$ is defined everywhere.)

$f'(c) = 0$ when $\sin c – \cos c = 0$: \begin{align*} \sin c – \cos c &= 0 \\ \\ \sin c &= \cos c \\ \\ \tan c &= 1 &&\text{[Recall that our interval is }[0, 2\pi]]\\ \\ c &= \frac{\pi}{4}, \, \frac{5\pi}{4} \quad \cmark \end{align*}

To find the global (absolute) maximum and minimum, we must compute the values of $f$ at the critical points ($\dfrac{\pi}{4}$ and $\dfrac{5\pi}{4}$) and at the endpoints (0 and $2\pi$):

$\bullet$ $f\left(\dfrac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \sin \left(\dfrac{\pi}{4}\right) = e^{-\frac{\pi}{4}} \dfrac{1}{\sqrt{2}} \approx 0.322$
$\bullet$ $f\left(\dfrac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \sin \left(\dfrac{5\pi}{4}\right) = e^{-\frac{5\pi}{4}} \left(-\dfrac{1}{\sqrt{2}}\right) \approx -0.0139$
$\bullet$ $f(0) = e^0 \sin (0) = 0$
$\bullet$ $f(2\pi) = e^{-2\pi} \sin (2\pi) = 0$

The global maximum therefore occurs at the point $\left(\dfrac{\pi}{4}, 0.322 \right)$, while the global minimum occurs at $\left(\dfrac{5\pi}{4}, -0.0139\right)$. $\quad \cmark$

For comparison purposes, the graph of the function is plotted below.

This question is asking the same thing as, “when is the particle’s position instantaneously not changing?”, which in turn is the same question as, “when is the derivative of $x(t)$ zero?”.

\begin{align*} x'(t) &= \frac{d}{dt} \left[ (t -2)^3 (t-6)\right] \\ \\ &= \left(\frac{d}{dt}(t -2)^3 \right) (t – 6) + (t-2)^3 \left( \frac{d}{dt}(t-6) \right) \\ \\ &= 3(t-2)^2 (t-6) + (t-2)^3(1) \\ \\ &= (t-2)^2\left[3(t-6) + (t-2) \right] \\ \\ &= (t-2)^2\left[ 3t -18 + t – 2 \right] \\ \\ &= (t-2)^2 \left[4t – 20 \right] \\ \\ &= 4 (t-2)^2 (t – 5) \end{align*} We want to find when $x'(t) = 0$:

\begin{align*} x'(t) &= 0 \\ \\ 4 (t-2)^2 (t – 5) &= 0 \\ \\ t &= 2, \, 5 \quad \cmark \end{align*}

This question is asking the same thing as, “when is the particle’s position increasing?”, which in turn is the same question as, “when is the derivative of $x(t)$ positive?”.

We determined in part (a) that $$x'(t) = 4 (t-2)^2 (t – 5)$$ We want to know when $x'(t) > 0$. Since $(t-2)^2$ is always positive, $x'(t)$ is positive when $(t – 5)$ is positive, or $$t > 5 \quad \cmark$$

We know that the particle could change directions at the times we found in part (a), when it is momentarily at rest: $t = 2, \, 5$. But to see if it actually does, we need to examine the direction it is moving during the three intervals (i) $t < 2$; (ii) $2 < t < 5$; and (iii) $t < 5$.

As usual, we compute the value of $x'(t)$ at some convenient values of $t$ for each of the three intervals:
$$\bullet\, x'(1) = 4(-1)^2(-4) = -16 \text{, so $x'(t)$ is negative for $t < 2$.}$$
$$\bullet\, x'(3) = 4(1)^2(-2) = -8 \text{, so $x'(t)$ is negative for $2 < t < 5$.}$$
$$\bullet\, x'(6) = 4(4)^2(1) = 64 \text{, so $x'(t)$ is positive for $5 < t$.}$$
So the particle changes direction only once, from moving to the left ($x'(t)$ < 0: its position decreases) to moving to the right ($x'(t) > 0$: its position increases) at $t = 5$. $\quad \cmark$

This question is asking the same thing as, “what is the minimum value of $x(t)$?”. We identified the critical points of $x(t)$ in part (a). In part (c), we found that at $t = 5$ the particle switches from moving to the left to moving to the right, and so that instant marks its most negative position.

Hence its position furthest to the left is $$x(5) = (3)^3(-1) = -27 \quad \cmark$$