Question

If the function $f$ be given by $f(x)=x^3-3x+3$, then
I. $x=\pm 2$ are the only critical points for local maxima or local minima.
II. $x=1$ is a point of local minima.
III. local minimum value is $2$ .
IV. local maximum value is $5$ .

• A. Only I and II are true
• B. Only II and III are true
• C. Only I, II and III are true
• D. Only II and IV are true

Answer 1

Answer: (D)

We have $f(x)=x^3-3x+3$
or $f^{\prime }(x)=3x^2-3=3(x-1)(x+1)$
or $f^{\prime }(x)=0$ at $x=1$ and $x=-1$
Thus, $x=\pm 1$ are the only critical points which could possibly be the points of local maxima and/or local minima of $f$. Let us first examine the point $x=1$.
Note that for values close to $1$ and to the right of $1.f$ ${}^{\prime }(x)>0$ and for values close to $1$ and to the left of $1.f$ ${}^{\prime }(x)<0.$
Therefore, by first derivative test, $x=1$ is a point of local minima and local minimum value is $f(1)=1.$
In the case of $x=-1$, note that $f^{\prime }(x)>0$, for values close to and to the left of $-1$ and $f^{\prime }(x)<0$, for values close to and to the right of $-1$.
Therefore, by first derivative test, $x=-1$ is a point of local maxima and local maximum value is $f(-1)=5$

Values of x		Sign of $f^{\prime }(x)=3(x-1)(x+1)$
Close to 1	to the right (say 1 .1 etc.)	> 0
	to the left (say 0.9 etc.)	< 0
Close to -1	to the right (say -0.9 etc.)	< 0
	to the left (say -1.1 etc.)	> 0