# How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=x^3+6x^2-36x#?

The function has a relative minimum at

The function has a relative maximum at

Given -

#y=x^3+6x^2-36x#

#dy/dx=3x^2+12x-36#

#(d^2y)/(dx^2)=6x+12#

#dy/dx=0 => 3x^2+12x-36=0#

#3x^2+12x-36=0# [Dividing both sides by 3 we get]

#x^2+4x-12=0#

#x^2+6x-2x-12=0#

#x(x+6)-2(x+6)=0#

#(x-2)(x+6)=0#

#x-2=0#

#x=2#

#x+6=0#

#x=-6#

At

#(d^2y)/(dx^2)=6(2)+12=12+12=24>0#

At

Hence the function has a relative minimum at

At

#(d^2y)/(dx^2)=6(-6)+12=-36+12=-24<0#

At

Hence the function has a relative maximum at

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To find the exact relative maximum and minimum of the polynomial function ( f(x) = x^3 + 6x^2 - 36x ), follow these steps:

- Find the derivative of the function: ( f'(x) = 3x^2 + 12x - 36 ).
- Set the derivative equal to zero and solve for ( x ) to find critical points: ( 3x^2 + 12x - 36 = 0 ).
- Solve the quadratic equation to find the critical points: ( x = -6 ) and ( x = 2 ).
- Evaluate the function at the critical points and at the endpoints of the interval of interest, if any.
- The maximum and minimum values correspond to the highest and lowest function values obtained in step 4, respectively.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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