Reading a curve from its derivatives
Tonight's idea
Take \(f(x) = x^3 - 3x\). Three questions tell you its shape.
Where is \(f'\) zero? \(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\), zero at \(x = \pm 1\). For \(|x| > 1\) the slope is positive, between them negative. So the curve rises, flattens at \((-1, 2)\), falls, flattens at \((1, -2)\), rises again. The first flat point is a local maximum, the second a local minimum.
Where does \(f''\) change sign? \(f''(x) = 6x\). For \(x \lt 0\) the slope is decreasing, so the curve bends downwards (concave down, like a hilltop); for \(x > 0\) it bends upwards (concave up, like a bowl). The switch, at \((0, 0)\), is a point of inflection. This also settles which flat point is which without checking slopes either side: \(f''(-1) = -6 \lt 0\) means hilltop, a maximum; \(f''(1) = 6 > 0\) means bowl, a minimum.
What happens far away? For large \(|x|\), \(x^3\) swamps \(3x\): the curve comes up from the bottom left and leaves at the top right. Roots, if easy, pin it further: \(x(x^2 - 3) = 0\) at \(x = 0, \pm\sqrt3\). Thread a smooth curve through everything you know, and you have the sketch.
Watch one: \(f(x) = x\,e^{-x}\)
Product rule: \(f'(x) = e^{-x} - x e^{-x} = (1 - x)e^{-x}\). Since \(e^{-x} > 0\) always, the sign is that of \(1 - x\): rising before \(x = 1\), falling after. Maximum at \((1, 1/e) \approx (1, 0.37)\).
Again: \(f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}\). Sign change at \(x = 2\): inflection at \((2, 2/e^2) \approx (2, 0.27)\), where the fall stops steepening.
Far away: as \(x \to \infty\), \(e^{-x}\) shrinks faster than \(x\) grows, so \(f \to 0\) from above, hugging the axis. As \(x \to -\infty\), both factors blow up and \(f \to -\infty\). Through the origin. Sketch: up from the bottom left, over a low hump at \(x = 1\), then a long slide down towards the axis.
Slide a graph up and down with the constant \(k\), the move the stretch relies on: mathdetours.com/demos/transform.html.
Your turn
- \(f(x) = x^3 - 6x^2 + 9x\). Find the turning points, the inflection point and the roots, and sketch. Which root does the curve touch rather than cross?
- \(f(x) = x^4 - 2x^2\). Find all three flat points and classify them with \(f''\). Then find the two inflection points. Sketch, using the symmetry \(f(-x) = f(x)\).
- \(f(x) = \dfrac{x}{1 + x^2}\). Show \(f'(x) = \dfrac{1 - x^2}{(1+x^2)^2}\), find the maximum and minimum, and say what happens as \(x \to \pm\infty\). Sketch. (The three inflection points are at \(0\) and \(\pm\sqrt3\); you need not derive them.)
- \(f(x) = \dfrac{\ln x}{x}\) for \(x > 0\). Find its one turning point and its behaviour as \(x \to 0^+\) and \(x \to \infty\), and sketch. Then use the sketch to decide, without a calculator, which is larger: \(e^{\pi}\) or \(\pi^{e}\).
Stretch: counting roots from the sketch
For which values of \(c\) does \(x^3 - 3x + c = 0\) have three distinct real roots? Use the sketch of \(x^3 - 3x\), not a formula. What happens at the boundary values of \(c\)?
Answers
- \(f' = 3x^2 - 12x + 9 = 3(x-1)(x-3)\): maximum \((1, 4)\), minimum \((3, 0)\). \(f'' = 6x - 12\): inflection \((2, 2)\). \(f = x(x-3)^2\), roots \(0\) and \(3\); the curve touches the axis at \(x = 3\) (the minimum sits on it) and crosses at \(0\). Far away like \(x^3\).
- \(f' = 4x^3 - 4x = 4x(x-1)(x+1)\): flat at \(x = -1, 0, 1\). \(f'' = 12x^2 - 4\): \(f''(\pm1) = 8 > 0\), minima at \((\pm1, -1)\); \(f''(0) = -4 \lt 0\), maximum at \((0, 0)\). Inflections where \(12x^2 = 4\), \(x = \pm 1/\sqrt3\), \(y = -5/9\). A W shape, symmetric, roots \(0, \pm\sqrt2\).
- Quotient rule: \(f' = \dfrac{(1+x^2) - x \cdot 2x}{(1+x^2)^2} = \dfrac{1 - x^2}{(1+x^2)^2}\). Maximum \((1, \tfrac12)\), minimum \((-1, -\tfrac12)\). As \(x \to \pm\infty\), \(f \to 0\) (denominator wins), so the axis is a horizontal asymptote. An odd function: an S through the origin, flattening out both ways.
- \(f' = \dfrac{1 - \ln x}{x^2}\), zero at \(x = e\): maximum \((e, 1/e)\). As \(x \to 0^+\), \(\ln x \to -\infty\) so \(f \to -\infty\); as \(x \to \infty\), \(f \to 0\) (the inflection is at \(x = e^{3/2}\)). The maximum value of \(\ln x / x\) is at \(x = e\), so \(\ln\pi/\pi \lt \ln e/e = 1/e\). Multiply by \(\pi e\): \(e\ln\pi \lt \pi\), so \(\pi^e \lt e^\pi\). (About 22.5 against 23.1.)
Stretch. Adding \(c\) slides the graph up by \(c\). The local maximum moves to \(2 + c\) and the local minimum to \(-2 + c\). Three crossings of the axis need the hump above the axis and the dip below it: \(2 + c > 0\) and \(-2 + c \lt 0\), so \(-2 \lt c \lt 2\). At \(c = 2\) the minimum touches the axis at \(x = 1\): a double root, \(x^3 - 3x + 2 = (x-1)^2(x+2)\). At \(c = -2\) the same at \(x = -1\). Beyond \(\pm2\), one real root.