cube root of (x^3-9x)
asked Jan 29, 2014 in Calculus Answers by Halimah (200 points)

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2 Answers

y=x^3-9x   or x*(x^2-9)

dy/dx=3x^2-9

min & max ...find weer dy/dx=0...find the zeroes av 3x^2-9

=3*(x^2 -3)

zeroes at x=+-sqrt(3)

at +sqrt(3), y=sqrt(3)*(3-9)..-.6*sqrt(3)

at -sqrt(3), y=-sqrt(3)*(3-9)...+6*sqrt(3)
answered Jan 29, 2014 by muneepenee

Question: find the maximum and minimum values of the function, cube root of (x^3-9x).

f = (x^3 - 9x)^(1/3)

f' = df/dx = (1/3)(x^3 - 9x)^(-2/3)*(3x^2 - 9)

To find turning points, set f' = 0.

(1/3)(x^3 - 9x)^(-2/3)*(3x^2 - 9) = 0

(x^3 - 9x)^(-2/3)*(3x^2 - 9) = 0

(3x^2 - 9) / (x^3 - 9x)^(2/3) = 0

i.e. 3x^2 - 9 = 0

x^2 = 3

x = +/- sqrt(3)

To see which of these values for x gives a maximum or minimum, get f'' and see whether it is positive or negarive at x = =/- sqrt(3).

f'' = -(2/9)*(3*x^2-9)^2/(x^3-9*x)^(5/3)+2*x/(x^3-9*x)^(2/3)

at x = -sqrt(3), f'' = -(1/3)*6^(1/3)*3^(1/6)

Since f'' is negative, then f(x) is at a maximum at x = -sqrt(3),

at x = sqrt(3), f'' = 2*sqrt(3)/(6*sqrt(3))^(2/3)

Since f'' is negative, then f(x) is at a minimum at x = sqrt(3),

f_min = f(sqrt(3)) = -6^(1/3)*3^(1/6)

f_max = f(-sqrt(3)) = 6^(1/3)*3^(1/6)

 

answered Jan 29, 2014 by Fermat Level 11 User (80,500 points)
find the maximum and minimum values of the functions

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