A Derivative That Cannot Jump: Darboux Versus a Piecewise Construction
A student is asked whether there exists a differentiable function f:(-1,1)→R with f′(x)=0 for x<0 and f′(x)=1 for x>0. The student proposes f(x)=0 for x≤0 and f(x)=x for x>0, and writes: 'On each side the derivative is correct. At 0 the left derivative is 0 and the right derivative is 1; since the problem only requires f′(0)=1, I define f′(0)=1.' The instructor marks this wrong. A second student suggests replacing the corner by a cubic on (-δ,δ) so that f is C^1 and the derivative transitions smoothly from 0 to 1. The problem asks you to decide whether any differentiable f can satisfy the stated derivative conditions, to identify the precise failure in the first construction, and to evaluate whether Darboux's theorem or the mean value theorem is the cleaner tool. A third student claims that because derivatives can be discontinuous, Darboux's theorem does not apply.
1. Show that no differentiable f can satisfy f′(x)=0 for x<0 and f′(x)=1 for x>0. Address the first construction and the third student's objection.
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On (-1,0), f′=0, so by the mean value theorem f is constant there; write f(x)=C for x<0. On (0,1), f′=1, so f(x)=x+D there. Continuity at 0 forces C=D, since both one-sided limits equal f(0). Hence f(x)=C for x≤0 and f(x)=x+C for x>0. Now compute the derivative at 0. For h<0, (f(h)−f(0))/h=(C−C)/h=0. For h>0, (f(h)−f(0))/h=(h+C−C)/h=1. The one-sided limits differ, so f′(0) does not exist. Thus f is not differentiable at 0, regardless of how f′(0) is defined. The first construction is exactly this with C=0; declaring f′(0)=1 does not make the derivative exist.
The third student is wrong: Darboux's theorem applies to every derivative, even a discontinuous one. Since f′(−1/2)=0 and f′(1/2)=1, Darboux would force some ξ with f′(ξ)=1/2, impossible under the stated values (and if ξ=0, differentiability at 0 already fails). The mean value theorem gives the cleanest direct contradiction; Darboux is also valid but needs the extra remark about ξ=0.
2. Can a cubic smoothing or a C^1 transition repair the construction? Distinguish allowing a transition interval from keeping the exact derivative conditions.
Read the worked answer
A C^1 transition can be built only if the derivative is allowed to change on an interval. For example, on a transition interval (−δ,δ) one can prescribe f′ to rise continuously from 0 to 1, using a smoothstep such as 3t^2−2t^3 after rescaling, and integrate it. That produces a C^1 function with f′=0 for x≤−δ and f′=1 for x≥δ. But such a function has 0<f′<1 at interior points of the transition, so it does not satisfy f′=0 for every x<0 and f′=1 for every x>0.
If the exact conditions are kept, no repair is possible. The derivative would have to take only the values 0 and 1 on the two sides, with no interval on which to interpolate. Differentiability at 0 forces the left and right difference quotients to agree, but they are 0 and 1. Equivalently, Darboux forbids a derivative from jumping over 1/2. The cubic idea changes the problem by inserting a transition region; it does not produce a differentiable function satisfying the original piecewise derivative specification.