Question

What is the total number of relative maximum and minimum points of the function whose derivative for all x is given by f ‘ (x) = x(x – 3)^2 (x + 1)^4 ?

Answer 1

Answer:

There are in total 3 maximum and minimum points of f(x)

Step-by-step explanation:

To find the number of stationary points (maximum and minimum points) in any function (f(x)), we need to find for what values of x does the derivative of f(x) equal to 0.

mathematically, for how many values of x does f'(x) = 0

$f'(x) = x(x-3)^2(x+1)^4$

$0 = x(x-3)^2(x+1)^4$

since the factors can be separately solved we can write

$0 = x, 0 = (x-3)^2, 0 = (x+1)^4$

$x= 0, \sqrt{(x-3)^2} = \sqrt{0}, \sqrt{(x+1)^4}=\sqrt{0}$

$x=0, (x-3)=0, (x+1)=0$

$x = 0, x = 3, x = -1$

Since there are 3 answers of x for which f'(x) = 0, we can say that the there are a total of 3 stationary (maximum or minimum) points in f(x).