# Why is there no local min for this graph?

Graph of $f'(x)$

Supposedly there is no local minimum on the interval (0, 8) for the above graph, but I don't understand why. I thought there were local mins at x = 1 and x = 6.

• The function being graphed has local minima, but your link suggests that this is the graph of the derivative of the function you are inspecting? – turkeyhundt Dec 15 '16 at 21:14
• You have the graph of $f'(x)$, while I suspect that the question asks about the minima of $f(x)$. Can that be right? – Arthur Dec 15 '16 at 21:14
• Oh! It is asking for the minima of f(x). Is the minima of f(x) where f'(x) is above 0? – user389369 Dec 15 '16 at 21:59

It sounds like the exercise is asking you to explain that there are no local minima of $f$, providing you with the graph of $f'$. If this interpretation of the exercise is correct, consider that candidates for local extrema of $f$ are at zeros of $f'$. There is one zero of $f'$ in the middle of the picture, but no sign change in $f'$ occurs there, so this is not a local extremum. (Near this point the graph would look something like the graph of $y=-x^3$ near $x=0$.)
• Not of the function $f$. – Brian M. Scott Dec 15 '16 at 21:14
• Since the OP has been told that whatever function is under consideration has no local minimum in $(0,8)$, and since we’re clearly given the graph of $f'$, not of $f$, it is easily inferred from the available information that the original question was almost certainly about $f$, not about $f'$. – Brian M. Scott Dec 15 '16 at 21:21