Find all of the critical points of the given function afx x

Find all of the critical points of the given function. a)f(x) = x^2/3 (x - 2)^2 b) g(x) = [x] c) h(x) = (x + 2)^2(2x - 3)^5 Find the maximum and minimum values of f(x) = sin^2x + cos x on [0,pi].

Solution

a) f(x) = x^2/3( x - 2 )²

We first need the derivative of the function in order to find the critical points and so let’s get that and notice that we’ll factor it as much as possible to make our life easier when we go to find the critical points.

f\'(x) = 4/3( x - 2 )(2x^2/3 - x^-1/3 )

Now, our derivative is a polynomial and so will exist everywhere. Therefore the only critical points will be those values of x which make the derivative zero. So, we must solve

4/3( x - 2 )(2x^2/3 - x^-1/3 ) = 0

Because this is the factored form of the derivative it’s pretty easy to identify the three critical points. They are

x = 0 , x = 2

b) g(x) = [x]

g\'(x) = ( x - h ) = x ,    { h < 0.5 }

   x = 0 , 1 , 2 , 3 , .........

   g\'(x) = ( x + h ) = x + h , { h >= 0.5 }

   if x + h = 1.61

  then x = 2

so critical point will be greater then x+ h and fixid value.

c) h(x) = ( x + 2 )²(2x - 3 )⁵

h\'(x) = 14( x+ 2).( 2x - 3)⁴.(x + 1)

14( x+ 2).( 2x - 3)⁴.(x + 1) = 0

critical points are

x = -2 , x = 3/2 , x = -1