When finding relative minimum/maximum, what is the point of using the second derivative test? Why use the second derivative test over the first derivative test when finding maxima and minima if it's uncertain what $f''(x) = 0$ is? Why not just always use the first derivative since we need to take the first derivate either way?

Let $f(x) = \sin(\mathrm{e}^{16-x^2})$ and determine whether $x = 0$ is a critical point, and if so, whether $f$ has a local minimum, local maximum, or neither at $x = 0$.

(The small voids near the upper and lower bounds are numerical artifacts.  A more accurate plot would look like a solid rectangle.)  I immediately notice a difficulty with the first derivative attack on the problem: the function oscillates vigorously near $x = 0$.
Then $f'(x) = -2x\mathrm{e}^{16-x^2}\cos(\mathrm{e}^{16-x^2})$.  We can immediately see $x = 0$ is a critical point.  The first derivative has a lot of zeroes near $x = 0$.

It's "easy enough" to solve for the zeroes of the derivative:  $x = 0$, $x = \pm \sqrt{16 + \ln(2/\pi)}$, $x = \pm \sqrt{16+\ln\left( \frac{-2}{\pi - 4 \pi k} \right)}$, or $x = \pm \sqrt{16+\ln\left( \frac{2}{\pi + 4 \pi k} \right)}$, for any integer $k$ with $1 \leq k \leq 1\,414\,268$.  The fourth case with $k = 1\,414\,268$ gives the critical points closest to (and distinct from) $x=0$: $x = \pm 0.00034073{\dots}$.  (How much work would one show to justify the claims in the previous sentence?)  But all of this analysis has been a stupendous waste of time.
Instead,
$$  f''(x) = (4x^2 - 2)\mathrm{e}^{16-x^2}\cos(\mathrm{e}^{16-x^2}) \\ - 4 x^2 \mathrm{e}^{32-2x^2}\sin(\mathrm{e}^{16-x^2})  $$ and $f''(0) = -2 \mathrm{e}^{16}\cos(\mathrm{e}^{16}) = 1.5251 \times 10^{7}$, so $f$ has a local minimum at $x = 0$.


We don't only run into problems with the "first derivative strategy" only when we use transcendental functions.  Consider
$$  f(x) = \frac{1}{6} x^6 - \frac{6}{5} x^5 - \frac{\mathrm{e}^2 + \pi^2 - 14}{4} x^4 + \frac{4}{3}(\mathrm{e}^2 + \pi ^2 - 4) x^3 + \frac{1}{2}(9 - 5\mathrm{e}^2  - 5\pi ^2 +\mathrm{e}^2\pi^2)x^2 - 2 (\mathrm{e}^2 -1)(\pi ^2 - 1) x + 7  \text{.}  $$
It's just a polynomial.  All its derivatives are polynomials, how bad could it be?
$$  f'(x) = x^5 - 6x^4 - (\mathrm{e}^2 + \pi^2 - 14)x^3 + 4(\mathrm{e}^2 + \pi^2 - 4)x^2 + (9 - 5\mathrm{e}^2 - 5 \pi^2 + \mathrm{e^2}\pi^2) x - 2 (\mathrm{e}^2 - 1)(\pi^2 - 1)  \text{.}  $$
If we abuse the rational root test, applying it even though the coefficients are not integers, we can find that $x = 2$ is a root of $f'$.
$$  f'(x) = (x-2)(x^4 - 4x^3 - (\mathrm{e}^2 + \pi^2 - 6)x^2 + (2\mathrm{e}^2 + 2\pi^2 - 4)x + \mathrm{e}^2\pi^2-\mathrm{e}^2 - \pi^2 + 1)  \text{.}  $$
Good luck factoring that.  (It's $f'(x) = (x-2)((x-1)^2 - \pi^2)((x-1)^2-\mathrm{e}^2)$.)
However, we can characterize the critical point at $x = 2$ immediately.
$$  f''(2) = (\pi^2 - 1)(\mathrm{e}^2 - 1) = 56.668{\dots}  \text{.}  $$
