Which are the local extremes of y = x^3 - 9x ?Which are the local extremes of y = x^3 - 9x ?

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should remember that you may find the extremes of the function solving the equation .

Differentiating with respect to x yields:

Solving the equation yields:

Hence, the function reaches its maximum at and it reaches its minimum at

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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y= x^3 - 9x

First we will differentiate"

y' = 3x^2 - 9

Now to determine the critical values we need to find the derivative's zeros.

==> 3x^2 - 9 = 0

==> 3x^2 = 9

==> x^2 = 3

==> x= +-sqrt3

Then we have two extreme values:

when x= sqrt3

==> y= (sqrt3)^3 - 9*(sqrt3) = 3sqrt(3) - 9sqrt3= -6sqrt3

when x= -sqrt3

==> y= (-sqrt3)^3 - 9*-sqrt3 = -3sqrt3 +9sqrt3 = 12sqrt3

Then y has two extreme values:

(sqrt3, -6sqrt3)   and (-sqrt3, 6sqrt3)

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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Local extremes occur for the values of x for which:

dy/dx = 0

given:

y = x^3 - 9x

Therefore:

dy/dx = 3x^2 - 9

Equating this to 0 we get:

3x^2 - 9 = 0

==> x^2 = 3

==> x = 3^(1/2)

Substituting this value of x in given equation we get extreme values of  as:

y = (3^1/2)^3 -  9*3^1/2

= 3^(3/2) - 3*3^(3/2)

= -2*3^(3/2)

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neela | High School Teacher | (Level 3) Valedictorian

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To determine local extremes of y = x^3-9x.

LOcal extremes  are given by solving  y' (x) = 0 .

y' = (x^3-9x)' = 3x^2-9

y'= 0 gives  3x^2-9 = 0. Or  x^2= 3.

So x = sqrt3 Or x = -sqrt 3

So y(sqrt3) = sqr3^(3/2) - 9(sqrt3) = 3 (sqrt3) - 9sqrt(3) =  -6sqrt3

y(-sqrt3) =x^3-9x =  3(-sqrt3) - 9(-sqrt3) = -3sqrt3 +9sqrt3 = 6sqrt3.

So at x = sqrt3 y = -6sqrt3 and at x = -sqrt3, y = 6sqrt3 are the local minimum and maximum  or extreme values.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the local extremes of a function, minimum or maximum, we'll do the first derivative test. 

y'= (x^3-9x)'

y'= 3x^2 - 9

Now, we'll calculate the roots of the first derivative. Each root of derivative represents the value for the function y has an extreme value.

3x^2 - 9 = 0

We'll factorize:

3(x^2-3)=0

We'll solve the difference of squares:

3(x-sqrt3)(x+sqrt3)=0

We'll put each factor of the product as zero.

x-sqrt3=0

We'll add sqrt3 both sides:

x=sqrt3

x+sqrt3=0

x=-sqrt3

So, the extreme values of the function are:

f(sqrt3) = (sqrt3)^3-9sqrt3

f(sqrt3) = 3sqrt3 - 9sqrt3

f(sqrt3) = -6sqrt3

f(-sqrt3) = -3sqrt3 + 9sqrt3

f(sqrt3) = 6sqrt3

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