Q: Find the critical numbers of function g(x) = (x^(2/3)) * ((x - 6)^2) by using the definition and doing the necessary algebra by hand.
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in order to find the critical points of a function, we need to find the first derivative and equal it to zero as the following:
g'(x) = x^(2/3) * 2(x - 6) * 1 + (x - 6)^2 * (2/3) * x^(-1/3)
g'(x) = (x - 6) * [ x^(2/3) * 2 + (x - 6) * (2/3) * x^(-1/3) ]
g'(x) = (x - 6) * x^(2/3) * [ 2 * x^(2/3 - 2/3) + (x - 6) * (2/3) * x^(-1/3 - 2/3) ]
g'(x) = (x - 6) * x^(2/3) * [ 2 * x^0 + (x - 6) * (2/3) * x^(-3/3) ]
g'(x) = (x - 6) * x^(2/3) * [ 2 * 1 + (x - 6) * (2/3) * x^-1 ]
g'(x) = (x - 6) * x^(2/3) * [ 2 + 2(x - 6) / (3x) ]
g'(x) = (x - 6) * x^(2/3) * [ 2*3x/3x + (2x - 12) / (3x) ]
g'(x) = (x - 6) * x^(2/3) * [ ( 6x + 2x - 12) / (3x) ]
g'(x) = (x - 6) * x^(2/3) * [ ( 8x - 12) / (3x) ]
x - 6 = 0 -----> x = 6 critical point
x^(2/3) = 0 -----> x = 0 critical point but we are going to not include it since :
[ ( 8x - 12) / (3x) ] <----- 0 is not differentible
8x - 12 = 0 -----> x = 3/2 ( critical point )
critical points are:
6, 3/2 & 0
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g'(x) = x^(2/3) * 2(x - 6) * 1 + (x - 6)^2 * (2/3) * x^(-1/3)
g'(x) = (x - 6) * [ x^(2/3) * 2 + (x - 6) * (2/3) * x^(-1/3) ]
g'(x) = (x - 6) * x^(2/3) * [ 2 * x^(2/3 - 2/3) + (x - 6) * (2/3) * x^(-1/3 - 2/3) ]
g'(x) = (x - 6) * x^(2/3) * [ 2 * x^0 + (x - 6) * (2/3) * x^(-3/3) ]
g'(x) = (x - 6) * x^(2/3) * [ 2 * 1 + (x - 6) * (2/3) * x^-1 ]
g'(x) = (x - 6) * x^(2/3) * [ 2 + 2(x - 6) / (3x) ]
g'(x) = (x - 6) * x^(2/3) * [ 2*3x/3x + (2x - 12) / (3x) ]
g'(x) = (x - 6) * x^(2/3) * [ ( 6x + 2x - 12) / (3x) ]
g'(x) = (x - 6) * x^(2/3) * [ ( 8x - 12) / (3x) ]
x - 6 = 0 -----> x = 6 critical point
x^(2/3) = 0 -----> x = 0 critical point but we are going to not include it since :
[ ( 8x - 12) / (3x) ] <----- 0 is not differentible
8x - 12 = 0 -----> x = 3/2 ( critical point )
critical points are:
6, 3/2 & 0
========
free to e-mail if have a question