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I solved this question below in a public contest and I don't know why one of the examiners gave me 12/15 points.

Question: If $a<b\in\mathbb R$ and $f:[a,b]\to \mathbb R$ a real function, continuous in $[a,b]$ and derivable in $(a,b)$. Show if $f'(x)=0$, for every $x\in (a,b)$, then there is $k\in \mathbb R$, such that $f(x)=k$, for every $x\in [a,b]$.

Solution: Let $h>0$ such that $a+h<b$. Applying the Mean Value Theorem in $[a+h,b]$, there is a $c\in(a+h,b)$ such that

$$f'(c)=\frac{f(b)-f(a+h)}{b-a-h}=0$$

It follows $f(a+h)=f(b)$ which also follows $f$ is constant $f(x)=k$ with $k=f(b)$.

The only mistake I found was forgetting to prove $f(a)=k$, are there other mistakes? I don't understand why I got only 80% of this question.

user42912
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    I couldn't say why a grader did this, but it is better to just say $x\in (a,b)$ rather than writing your arbitrary point as $a+h$. variable, which is confusing. But yes, skipping $h=0$ is a likely cause for the lost points. – Thomas Andrews Nov 02 '17 at 18:01
  • Yeah you should have just taken $\lim_{h \to 0^+}$ and then I think you'd have gotten full points for sure. – Mr. Chip Nov 02 '17 at 18:02
  • I might agree. You need to get $f(a)=k$ as well since your argument assumed $h >0$. This is a continuity argument, and is likely worth 20% of the problem. – Randall Nov 02 '17 at 18:09
  • Question has too many $x$s in it – DanielV Nov 02 '17 at 18:09
  • You mean too many commas? – Randall Nov 02 '17 at 18:15

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