critical point, increasing/decreasing function and local extremum

Question

If $f ' (x) =(x-1)^2 (x+2)$ is given and we're asked to find the 3 things in the title:

for critical points we put $f ' (x)=0$ and the values of x we get are critical points

for increasing/decreasing functions we will divide intervals according to critical point and take a test point to see sign of x and thus conclude if it's increasing or decreasing

for local extremums we can use second derivative test?

Are these three approaches correct or am i doing anything wrong?

Answer 1

Yes, you are right!

The first derivative test uses the derivatives of a function to locate the critical points of a function. To calculate the critical points we need to solve the equation $f^{'}(x)=0$. The computed values of the parameter $x$ will be critical.

The second-derivative test uses the value of the second derivative at those points to determine whether such points are a local maximum or a local minimum. If the function f is twice-differentiable at a critical point x (i.e. a point where $f^{'}(x)=0$), then:

1. If $f^{''}(x)<0$, then f has a local maximum at x.

2. If $f^{''}(x)>0$, then f has a local minimum at x.

3. If $f^{''}(x)=0$, the test is inconclusive.

Good luck!

Answer 2

Some corrections, for technical correctness:

for critical points we put $f ' (x)=0$

1. Critical points are where $f'(x)$
   • either doesn't exist $(y=|x|$ at $0)$
   • or equals $0.$

for local extremums we can use second derivative test

2. For local extrema that are stationary turning points, the sign test is one alternative (among others) to the second-derivative test.

3. But a local extremum might also be

   • a non-stationary turning point $(y=|x|$ at $0),$ or
   • a stationary non-turning point $(y=3$ at $0),$ or
   • neither stationary nor turning $(y=\sqrt x$ at $0)$ !

Answer 3

If you are going to divide the interval according to whether $f$ is increasing/decreasing anyway, then you don't need the second derivative test. Just look at the monotonicity of $f$ near its critical points to decide local extremums.

In your example, $x=-2$ is a critical point of $f$. Note that $f'(x) < 0$ for $x < -2$ and $f'(x) > 0$ for $-2 < x < 1$. This means $f$ is strictly decreasing on the left of $x=-2$ and strictly increasing on the right of $x=-2$. Thus, $x=-2$ is a strict local minimum of $f$.

You can do a similar analysis for another critical point $x=1$.