Let f:R→R be a differentiable function, and suppose f=f′ and f(0)=1. Then prove f(x)≠0 for all x∈R

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etsahalen5tt · Accepted Answer

Step 1Similar to the work of yearning4pi on this topic:With1)&amp;nbsp;f′(x)=f(x),&amp;nbsp;f(0)=1,we consider the function&amp;nbsp;e−xf(x); we have2)&amp;nbsp;(e−xf(x))′=−e−xf(x)+e−xf′(x)=e−x(−f(x)+f′(x))=0in light of (1); thus3)&amp;nbsp;e−xf(x)=c,;a constant,whence4)&amp;nbsp;f(x)=cex;then5)&amp;nbsp;c=ce0=f(0)=1,so that6)&amp;nbsp;f(x)=ex;clearly7)&amp;nbsp;ex=∑0∞dxnn!&amp;gt;0&amp;nbsp;for&amp;nbsp;x≥0,and if&amp;nbsp;x&amp;lt;0,&amp;nbsp;−x&amp;gt;0, so8)&amp;nbsp;e−x&amp;gt;0;but9)&amp;nbsp;e−xex=e−x+x=e0=1Long→ex=d1e−x&amp;gt;0;thus10)&amp;nbsp;∀x∈BR,;f(x)=ex&amp;gt;0.And then, of course, we always have Picard-Lindeloef as invoked by our colleague Severin Schraven.

Cailyn Hanson · Accepted Answer

Step 1There is a shorter proof if you know the Picard-Lindelöf theorem. Assume that there exists x0∈R such that f(x0)=0. Then f solves the ODE{f′=ff(x0)=0.Note that g≡0 also solves the ODE above, but by the uniqueness part of Picard-Lindelöf we would have f=g which contradicts f(0)=1.

Let \(\displaystyle{f}:{\mathbb{{R}}}\rightarrow{\mathbb{{R}}}\) be a differentiable function, and

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