Families of functions

What is it about?

A family of functions such as $f_k(x) = x^3 - kx$ contains one graph for each $k$ . When differentiating, $k$ is a constant: $f_k'(x) = 3x^2 - k$ . For $k > 0$ there are extrema at $x = \pm\sqrt{k/3}$ , for $k = 0$ a saddle point (no extremum), for $k < 0$ none. Solving $x^2 = k/3$ for $k$ ( $k = 3x^2$ ) and substituting gives the locus $y = -2x^3$ ( $x \neq 0$ ). All members pass through $(0 \mid 0)$ : $(k_2 - k_1) \cdot x = 0$ gives $x = 0$ . For $f_k(x) = (x - k) \cdot e^x$ there is no common point, since $(k_2 - k_1)e^x \neq 0$ . For $x \cdot e^{-kx}$ and $kx^2 - x^4$ all curves touch at the origin. Points of inflection move on a locus as well – for $x \cdot e^{-kx}$ on $y = \frac{x}{e^2}$ .

The extrema are red beads: when $k$ changes they glide along their locus and leave a glowing trace. The k-axis below the graph shows the case analysis; ▶ makes $k$ run.

Try it

x³ − kx: drag the maximum, then press “Run through k”.
Derivation from the start: press “Derivation: next step” several times.
Parabolas and x · e^(−kx): tap faint curves to jump to other values of $k$ .
Points of inflection: run through $k$ and compare both traces.
(x − k) · eˣ: why is there no common point?

Tasks

Task 1: Locus

Load task

Find the extrema of $f_k(x) = x^2 - 2kx + k$ and their locus.

Show solution

$f_k'(x) = 2x - 2k = 0 \Rightarrow \text{Min}(k \mid k - k^2)$ ; substituting $k = x$ gives the locus $y = x - x^2$ . Show

Task 2: Locus of x³ − kx

Load task

Find the extrema of $f_k(x) = x^3 - kx$ ( $k > 0$ ) and their locus; check with $k = 3$ .

Show solution

$x_E = \pm\sqrt{k/3}$ , so $k = 3x^2$ and $y = x^3 - 3x^2 \cdot x = -2x^3$ ( $x \neq 0$ ). For $k = 3$ : $\text{Max}(-1 \mid 2)$ , $\text{Min}(1 \mid -2)$ . Show

Task 3: No common point

Load task

For $f_k(x) = (x - k) \cdot e^x$ find the minimum, its locus and show that there is no common point.

Show solution

$f_k'(x) = e^x(x - k + 1) = 0 \Rightarrow \text{Min}(k - 1 \mid -e^{k-1})$ ; locus $y = -e^x$ . $(k_2 - k_1)e^x = 0$ is impossible for $k_1 \neq k_2$ . Show

Ideas for the classroom

Let students guess the locus from the glowing trace before computing it, then reveal the derivation step by step.
Mark the cases on a number line first and compare with the k-axis of the simulation.
Common stumbling blocks: treating $k$ as a variable when differentiating; forgetting the case analysis; dropping $k$ instead of solving for it; dividing by $k_1 - k_2$ without noting $k_1 \neq k_2$ .