For $f(x)=x^3-3x+3$, the point of local maxima and the local maximum value are
A$x=-1$, value $5$
B$x=1$, value $1$
C$x=-1$, value $1$
D$x=1$, value $5$
Answer & Solution
Correct answer: A. $x=-1$, value $5$
1. $f'(x)=3x^2-3=3(x-1)(x+1)$.
2. Critical points: $x=\pm 1$.
3. At $x=-1$: $f'$ changes $+$ to $-$, so local maxima.
4. $f(-1)=(-1)^3-3(-1)+3=-1+3+3=5$.
5. (At $x=1$ it is a local minimum with value $1$, not a maximum.)
_Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.18_
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