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$f=x^3-6x^2+12$
Suppose we restrict $f$ to the interval $[−1,6]$. Find all points in this interval at which $f$ has a local minimum or maximum.

My answer: It has a local minimum at: $x=4$ and local maximumn at: $x=0$

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    Does whatever text you're using allow local min/max at an endpoint of the interval of definition? – 2017-02-10
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    That's what confused me completely since *local* doesn't require intervals! – 2017-02-10
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    As well, I submitted the answer and it said "incorrect", the question is in TuDelft pre-univertisty calculus exam at EDX. – 2017-02-10
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    I'd try looking at the graph... – 2017-02-10
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    additionally you must compare the values $$f(-1),f(6),f(0)$$ and $$f(4)$$ – 2017-02-10

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The derivative of $f(x)= x^3- 6x^2+ 12$ is $f'(x)= 3x^2- 12$ which is 0 when x is 2 or -2. -2 is not in the given interval. The second derivative, $6x$, is positive at x= 2 so there is a local minimum there. Further the derivative is $3- 12= -9$ at x= -1. That is, f is decreasing there so values of f, for x> -1 are smaller- x= -1 is local maximum (relative to values in the given interval, x> -1). Finally, the derivative is $108- 12= 96$ at x= 6. f is increasing there so values of f, for x< 6, are smaller. x= 6 is a local maximum (relative to values in the given interval, x< 6). There is a local minimum at x= 2, a local maximum at x= -1, and a local maximum at x= 6.