# Answer to Question #8545 in Calculus for joe

Question #8545

given the curve y= 1/3 x3 (cubed) - 9x - 7

a) evaluate dy/.dx

b)for what values of x is dy/dx = 0

c)hence, state the values of x at which the curve possesses stationary points. (note not required to evaluate corresponding y values)

d)determine the nature of the stationary points of the curve.

a) evaluate dy/.dx

b)for what values of x is dy/dx = 0

c)hence, state the values of x at which the curve possesses stationary points. (note not required to evaluate corresponding y values)

d)determine the nature of the stationary points of the curve.

Expert's answer

given the curve y= 1/3 x³ - 9x - 7

&

a) evaluate dy/dx

dy/dx = (1/3)*3 x² - 9 - 0 = x² - 9.

b)for what values of x is dy/dx = 0

x² - 9 = 0

(x-3)(x+3) = 0

x = ±3 - zeros of dy/dx.

&

c)hence, state the values of x at which the curve possesses stationary points. (note not required to evaluate corresponding y values)

x = ±3 - stationary points.

&

d)determine the nature of the stationary points of the curve.

dy/dx > 0, x<-3;

dy/dx < 0, -3<x<3;

dy/dx > 0, x>3.

So, x=3 and x=-3 are local maximum and minimum points respectively.

&

a) evaluate dy/dx

dy/dx = (1/3)*3 x² - 9 - 0 = x² - 9.

b)for what values of x is dy/dx = 0

x² - 9 = 0

(x-3)(x+3) = 0

x = ±3 - zeros of dy/dx.

&

c)hence, state the values of x at which the curve possesses stationary points. (note not required to evaluate corresponding y values)

x = ±3 - stationary points.

&

d)determine the nature of the stationary points of the curve.

dy/dx > 0, x<-3;

dy/dx < 0, -3<x<3;

dy/dx > 0, x>3.

So, x=3 and x=-3 are local maximum and minimum points respectively.

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