Derivatives have the intermediate value property even when discontinuous — f′(x)=2xsin(1/x)−cos(1/x) oscillates wildly near 0 yet steps on every level. Jump discontinuities are impossible in a derivative.
Prove Darboux's theorem: if f is differentiable on [a,b], then f′ takes every value between f′(a) and f′(b). (Hint: use the Extreme Value Theorem scene and Fermat's interior-extremum lemma.)
g(x)=f(x)−cx,g′(a)<0<g′(b)⟹∃x0∈(a,b):g′(x0)=0
g′(x0)=0⟹f′(x0)=c
Zoom 70 (window [±0.004]), test level y=0.4: the discontinuous f′ meets the horizontal line 704 times in this window (at grid resolution) — the Darboux property in action.Show proof
First Fermat's lemma: if g has a local max (or min) at an interior point x0∈(a,b) and g′(x0) exists, then g′(x0)=0. For a local min, hg(x0+h)−g(x0)≥0 for small h>0 and ≤0 for h<0; the two-sided limit defining the derivative is forced to 0. A local max is the same with signs flipped.
Assume f′(a)<f′(b) (apply the argument to −f if >; the equal case is trivial). Fix c between them and put g(x)=f(x)−cx: then g is differentiable, hence continuous, on [a,b], and g′(x)=f′(x)−c gives g′(a)<0<g′(b). By the Extreme Value Theorem (the EVT scene), g attains its minimum at some x0∈[a,b].
The point x0 is interior: since g′(a)<0, hg(a+h)−g(a)→g′(a)<0, so g(a+h)<g(a) for small h>0 — a is not the minimizer. Likewise g′(b)>0 gives g(b−h)<g(b) for small h>0 — b is not the minimizer either. Hence x0∈(a,b).
Applying Fermat's lemma at the interior minimizer x0 gives g′(x0)=0, hence f′(x0)=c. Since c was arbitrary between f′(a) and f′(b), f′ has the intermediate value property — even though it may be discontinuous.
Prove that f(x)=x2sin(1/x) (x=0), f(0)=0 is differentiable everywhere, but its derivative f′ is discontinuous at 0. (Get f′(0) from the definition and f′(x) for x=0 by the product and chain rules.)
Zoom 100 (window [±0.0005]), y=0: no amount of zooming calms the −cos(1/x) oscillation — 1579 crossings of f′ with the level in this window (at grid resolution).Show proof
Get f′(0) from the definition: hf(h)−f(0)=hh2sin(1/h)=hsin(1/h), and ∣hsin(1/h)∣≤∣h∣, so the squeeze theorem gives the limit 0. The numbers agree: ≈−0.00506 at h=0.01 and ≈−3.5×10−7 at h=10−6. Hence f′(0)=0.
For x=0, the product and chain rules give f′(x)=2xsin(1/x)+x2cos(1/x)⋅(−1/x2)=2xsin(1/x)−cos(1/x). So f is differentiable at every point, including 0.
But limx→0f′(x) does not exist. On ak=2kπ1→0 we have sin(1/ak)=0 and cos(1/ak)=1, so f′(ak)=−1; on bk=(2k+1)π1→0 we have sin(1/bk)=0 and cos(1/bk)=−1, so f′(bk)=1 (e.g. exactly −1.000000 and 1.000000 for k=5,10,20). If the limit existed it would have to be both −1 and 1 — a contradiction.
Since f′(0)=0 but limx→0f′(x) does not exist, f′ is discontinuous at 0 — the derivative of a differentiable function need not be continuous. Yet by Darboux's theorem (problem 1), this discontinuous f′ still has the intermediate value property.
Prove as a corollary of Darboux's theorem: no function can have the Heaviside step function H (0 for x<0, 1 for x≥0) as its derivative.
Zoom 0 (window [±0.5]), y=0.5: even the wildly oscillating f′ passes this level 46 times (at grid resolution) — a derivative's discontinuity can only be oscillation, never a jump.Show proof
Suppose some F, differentiable on all of R, had H as its derivative. Take a=−1, b=1: then F′(−1)=H(−1)=0 and F′(1)=H(1)=1. Since 0<21<1, Darboux's theorem (problem 1) gives some ξ∈(−1,1) with F′(ξ)=21.
But H takes only the values 0 and 1, so F′(ξ)=H(ξ)=21 is impossible — a contradiction. Hence no function has H as its derivative.
In general: a derivative cannot have a jump discontinuity. At a point where the one-sided limits exist but differ, Darboux forces the derivative to take the values strictly between them — which a step function never does. A derivative's discontinuities can only be essential oscillation, like the −cos(1/x) of problem 2.