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If $f$ is a function defined on $[0, 1]$ for which $f'(\frac{1}{2})$
exists but$f'(\frac{1}{2})$ is not an element in $[0,1]$

then f is discontinuous at
$x=\frac{1}{2}$. I need to find the statment is true or false and my conclusions are

  • the statement is true since ,if an inverse function exits and the domain of the inverse fuction should be the range of the function,in this case it fails. Is my analogy correct please help?
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up vote 4 down vote accepted

Nothing in the problem says anything about $f$ having an inverse. Consider the function $f(x)=3x$. What is $f'\left(\frac12\right)$? Is $f$ continuous on $[0,1]$?

Added: In fact, a function $f$ that is defined on an interval $[a,b]$ and has a derivative at a point $c\in(a,b)$ is always continuous at $c$. This is a standard result that is mentioned in just about all elementary calculus texts and proved in some.

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here $f'(x)=3$ and how can find $f'(\frac{1}{2})$ is it 3 ? since $f'(x)$ is constant and since the function is $y=mx$ its is continious at every point am i right? so i think $f$ is continious at $[0,1]$ – Biju jose Aug 16 '12 at 7:52
@Bijujose: Yes, that’s right: $f'\left(\frac12\right)=3$, which is not in $[0,1]$, but the function $f$ is continuous on all of $[0,1]$. In particular, it’s continuous at $1/2$. – Brian M. Scott Aug 16 '12 at 7:54
thanks Brian for helping ,i think i am a bit lack in elementary calculus ,so i need to make a concrete foundation in the basics,and also thanks for pointing out the theorem mentioned above – Biju jose Aug 16 '12 at 8:39

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