Curve sketching and connecting f, f' and f''

Day 1: Potira’s quartic

Potira wants an accurate sketch of f(x) = x^4 - 8x^3 + 18x^2.

A polynomial has every real number as its domain and no asymptotes, and the x^4 term sends f(x) \to \infty as x \to \pm\infty. Factoring, f(x) = x^2(x^2 - 8x + 18), and the quadratic has discriminant 64 - 72 < 0, so it is always positive: (0, 0) is the only intercept and f(x) > 0 for x \ne 0. There is no symmetry, since f(-1) = 27 while f(1) = 11.

The derivatives are

The factor 4x changes sign at 0 and (x - 3)^2 is never negative; 12(x - 1)(x - 3) is negative only between its zeros. So the charts, split at 0, 1 and 3, are:

f falls to a local minimum f(0) = 0 and rises from then on. At 3 the sign of f' does not change, so the point there, with f(3) = 81 - 216 + 162 = 27, is not an extremum: it is a level tangent in a rising graph. The concavity changes at 1 and 3, giving inflection points (1, 11), since f(1) = 1 - 8 + 18 = 11, and (3, 27).

From left to right (Figure 1): falling, concave up, into the valley; rising more and more steeply to (1, 11); rising less and less steeply to the flat point (3, 27); then rising more and more steeply. The zeros alone show only a touch at 0; the step at (3, 27) appears only through f' and f''.

Day 1: Kauan’s rational function

Kauan sketches g(x) = \dfrac{x}{(x - 2)^2}.

The domain is x \ne 2, the only intercept is (0, 0), and there is no symmetry (g(1) = 1 but \displaystyle g(-1) = -\frac{1}{9}).

As x \to 2 from either side the numerator tends to 2 and the denominator to 0 through positive values, so \lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = \infty: x = 2 is a vertical asymptote. The denominator has the higher degree, so \lim_{x \to \pm\infty} g(x) = 0 and y = 0 is a horizontal asymptote, approached from below on the left, where g < 0, and from above on the right.

Writing g(x) = x(x - 2)^{-2}, the product rule gives

and differentiating -(x + 2)(x - 2)^{-3} the same way,

g' breaks at -2 and at the gap 2; g'' breaks at -4 and 2, and its denominator is positive everywhere else, so g'' has the sign of x + 4.

So g falls to a local minimum \displaystyle g(-2) = \frac{-2}{16} = -\frac{1}{8}, climbs through the origin (with slope \displaystyle g'(0) = \frac{1}{4}) up the asymptote, and on the far side falls from \infty toward 0. It is concave down only on (-\infty, -4), so there is one inflection point, \displaystyle \left(-4, -\frac{1}{9}\right), since \displaystyle g(-4) = \frac{-4}{36}. The sign of g' also changes at 2, but 2 is not in the domain: no extremum there.

The dip is shallow (Figure 2). Zeros and asymptotes say only that the graph is negative left of 0; the derivatives find where it turns and bends.

Day 1: Araci’s logarithm

Araci sketches h(x) = x(\ln x)^2.

The domain is (0, \infty), so symmetry cannot occur. Since h(x) \ge 0, with h(x) = 0 only when \ln x = 0, the only intercept is (1, 0). At the open end, \lim_{x \to 0^+} x(\ln x)^2 = 0, because a positive power of x beats any power of \ln x (Comparing growth rates at infinity): the graph starts at the origin, which is not on it, and there is no vertical asymptote. As x \to \infty both factors grow, so there is no horizontal asymptote either.

By the product and chain rules,

h'(x) = 0 when \ln x = 0 or \ln x = -2, at x = 1 and x = e^{-2}. Both factors of h' are negative for x < e^{-2}, they have opposite signs between e^{-2} and 1, and both are positive beyond 1. h'' has the sign of \ln x + 1, which changes at x = e^{-1}.

So h rises to a local maximum h(e^{-2}) = e^{-2}(-2)^2 = 4e^{-2} \approx 0.541, falls to a local minimum h(1) = 0, and rises without bound. The concavity changes at e^{-1}, giving the inflection point (e^{-1}, e^{-1}), about (0.368, 0.368). One more limit shapes the start: h'(x) \to \infty as x \to 0^+, because (\ln x)^2 outgrows 2\ln x, so the graph leaves the origin almost vertically (Figure 3).

Day 2: reading f from the graph of f’

The top of Figure 4 is the graph of f' on [-2.4, 1.9]. It crosses the x-axis at -2, touches it at 1 without crossing, and has a high point at (-1, 4) and a low point at (1, 0). Where does f rise and fall, and where are its extrema and inflection points? Sketch f given f(-2) = 0.

Above or below the axis first. f' < 0 on (-2.4, -2), and f' > 0 on (-2, 1.9) except at the single point 1. So f is decreasing on [-2.4, -2] and increasing on [-2, 1.9]; one isolated zero of f' does not stop f from increasing. By the first derivative test, f has a local minimum at -2 and no extremum at 1, where f' touches the axis but keeps its sign.

Then rising or falling. The graph of f' rises on (-2.4, -1), falls on (-1, 1) and rises on (1, 1.9), so f'' > 0, then f'' < 0, then f'' > 0: f is concave up, down, up. The concavity changes at -1 and 1, the high and low points of the graph of f', and these are the inflection points of f. At 1 also f'(1) = 0, so that inflection point has a level tangent, like (3, 27) in Example 1.

The graph of f' gives no heights. With f(-2) = 0 the lowest point is (-2, 0), and the bottom of Figure 4 shows the resulting f, which is \displaystyle f(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 + 2x + 6 (its derivative is x^3 - 3x + 2 = (x - 1)^2(x + 2)). Any other function with this derivative is this one shifted up or down.

Day 2: reading f from the graph of f’’

Figure 5 is the graph of f'' on [-1.5, 3]; it touches the axis at 0 and crosses it at 2. Also f'(0) = 0 and \displaystyle f'(3) = \frac{9}{8}. Find the concavity and inflection points of f, and classify its critical points.

The sign of f'' gives the concavity directly. f'' < 0 on (-1.5, 0) and on (0, 2), and f'' > 0 on (2, 3). A single point where f'' = 0 does not stop f' from decreasing, so f is concave down on (-1.5, 2) and concave up on (2, 3), and the only inflection point is at x = 2. At 0 the graph of f'' touches the axis without crossing, so the concavity does not change there.

The rises and falls of f need f', and the graph of f'' gives only its direction: f' is decreasing on [-1.5, 2] and increasing on [2, 3]. Since f'(0) = 0 and f' is decreasing up to 2, f' > 0 on [-1.5, 0) and f' < 0 on (0, 2]. So f has a local maximum at 0. The second derivative test could not say this, because f''(0) = 0; the first derivative test does.

On [2, 3], f' increases from the negative value f'(2) to \displaystyle f'(3) = \frac{9}{8} > 0. It is continuous, being differentiable, so by the intermediate value theorem it is zero somewhere in (2, 3), and because it is increasing it is zero only once, at some c. There f' changes from negative to positive: a local minimum. The graph cannot place c exactly; for the curve drawn, \displaystyle f''(x) = \frac{1}{2}x^2(x - 2) with \displaystyle f'(x) = \frac{1}{24}x^3(3x - 8), it is \displaystyle c = \frac{8}{3}.

Day 2: matching f, f’ and f’’

Figure 6 shows a function f, its derivative f' and its second derivative f'', in some order, labeled A, B and C. Which is which?

C is positive everywhere. If C were the derivative of A or of B, that function would be increasing everywhere, with no turning point; but A and B both turn. So C is the derivative of neither, and C is f.

Then f' is A or B, and it must be zero where C turns (at 0), positive where C rises (x < 0) and negative where C falls (x > 0). A does all three. B is negative at 0, so it is not f'. Hence A is f' and B is f''. The last pair should agree, and it does: A turns at about \pm 0.58, where B crosses the axis; A falls between those points, where B is negative, and rises outside them, where B is positive. The graphs were drawn from f(x) = \dfrac{1}{1 + x^2}.

Day 2: a sketch from signs alone

No formula is given, only this: f is continuous everywhere, f(-1) = 3, f(0) = 1, f(1) = -1, \lim_{x \to -\infty} f(x) = -\infty, \lim_{x \to \infty} f(x) = 2, and

with f'(-1) = 0, f''(0) = 0, and neither f'(1) nor f''(1) existing. Sketch f and name its extrema and inflection points.

Plot the three points and the asymptote, then join them with the arcs the table names. Coming up from -\infty, f rises less and less steeply to (-1, 3), where f' changes from + to -: a local maximum with a level tangent. It falls more and more steeply, still concave down, to (0, 1), where the concavity changes: an inflection point. On (0, 1) it falls less and less steeply to (1, -1). There f' changes from - to + and f is continuous, so the first derivative test gives a local minimum although f'(1) does not exist: the graph has a corner. Beyond 1 it rises less and less steeply toward y = 2, from below, since it is increasing.

The concavity also changes at 1, from up to down, and f is continuous there, so (1, -1) is an inflection point as well as a minimum; the definition asks nothing of f' or f'' at the point. Figure 7 shows one function that fits, x^3 - 3x + 1 for x \le 1 and 2 - 3e^{-(x - 1)} for x > 1. Any sketch with the same points, arcs and asymptote is equally correct.