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?
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$\begingroup$Here's what I tell my students:
The first derivative test is usually harder to use, but will always give you an answer (at least for reasonable functions);
The second derivative test is usually easier to use, but it may not be applicable (if it is not a stationary point) and it may be inconclusive.
Why do I say that? To use the second derivative test, you just need to plug in a number and evaluate. That's straightforward. To use the first derivative test, you need to analyse the sign of the first derivative on a neighborhood of the point and figure out whether it changes signs as it goes through the point. This can be harder to do than just plugging in a single point.
As you say, you always already have the first derivative handy. But you may also need to have the second derivative handy: for example, if you are doing a full analysis of the function, looking for intervals in which the function is increasing and decreasing, and intervals in which the function is concave up and concave down, you will also have the second derivative handy.
It's a tradeoff, then:
For the First Derivative Test, you can always use it (stationary points and points where the first derivative does not exist), will almost always give you an answer either way (detecting critical points that are not relative extremes), and does not require any further derivatives to be calculated. But it requires a sign analysis, which may not be trivial (or possible).
For the Second Derivative, you can't always use it, and even if you can use it, it may be inconclusive and it will never detect critical points that are not relative extremes, and it requires you to calculate another derivative. But it is usually much easier to actually use, as it only requires figuring out the sign at a single point.
But more generally: it's always good to have more than one way to do something, as different ways may provide strengths and weaknesses that the other methods don't have, or provide a different kind of insight.
$\endgroup$ 4 $\begingroup$In most cases there is no point in making the second derivative test. Given a differentiable function $f: \>[a,b]\to{\mathbb R}$, a point $x\in\>]a,b[\>$ where $f'(x)\ne0$ cannot be an extremal point. Therefore you make a (hopefully finite) candidate list containing $a$, $b$, and the zeros $x_i$ $(1\leq i\leq r)$ of $f'$ in $\>]a,b[\>$. Then find the minimum or maximum function value for the points in this list.
It is different when you are not specifically out for extrema, but have a special point $\xi$ that interests you. Especially in a multivariate situation you might want to know whether this point $\xi$ is a local extremum, or what kind of saddle point. In such a case you of course have to look at the second derivatives (the Hessian).
$\endgroup$ $\begingroup$The second derivative test only works for local extrema. If you want global extrema, you have two options.
- If the function is defined on a closed bounded interval like $[-1,3]$. Then you can find the critical points, and compare the value at the critical points and on the side of the interval.
- Otherwise, you do need to know the variations of the function (i.e. the sign of the derivative). This also works for a closed bounded interval, and is often barely even more complicated than Method 1.
The thing is the second derivative test is essentially useless for single-variable calculus, since you can always almost always find the sign of the derivative and conclude.
The second thing is, you cannot speak of increasing / decreasing in multivariable calculus, and that is when the second derivative test really makes sense.
$\endgroup$ $\begingroup$Well $f'(x)=0$ means that the slope of the function is $0$ at point $x$. But you don't know if it is a maximum or a minimum or an inflection point. Take $x^2,-x^2$ and $x^3$ for example. The derivative of each function is 0 in 0, but the first one has a minimum, the next one maximum and the third one an inflection point in 0.
$\endgroup$ 3 $\begingroup$If you have a maximum or a minimum at $x$ then $f´(x) = 0$. But that does not make it true that if $f´(x) = 0$ you have a maximum or minimum at x. In order to know that you have a maximum/minimum at x, you need to know that both $f´(x)=0$ and $f´´(x) \neq 0$. Otherwise a function given by $f(x) = x^3$ would have a maximum or minimum at 0, because $3x^2(0) = 0$.
$\endgroup$ 2 $\begingroup$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.$$ f''(2) = (\pi^2 - 1)(\mathrm{e}^2 - 1) = 56.668{\dots} \text{.} $$
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