How do you locate the absolute extrema of the function #f(x)=x^312x# on the closed interval [0,4]?
The absolute extrema of a function,
So consider:
The minimum and maximum must occur at one of the values
To understand what we've done it may help to see the graph:
graph{y=(x^312x)*sqrt(4(x2)^2)/(sqrt(4(x2)^2)) [52.2, 40.28, 23.87, 22.4]}
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To locate the absolute extrema of the function ( f(x) = x^3  12x ) on the closed interval ([0,4]), follow these steps:
 Find the critical points of the function by setting the derivative equal to zero and solving for ( x ).
 Evaluate the function at the critical points and at the endpoints of the interval.
 The maximum and minimum values among these critical points and endpoints will be the absolute extrema.
Let's proceed with these steps:

Find the derivative of the function: ( f'(x) = 3x^2  12 ).

Set the derivative equal to zero and solve for ( x ): ( 3x^2  12 = 0 ). Solving this equation gives ( x = \pm 2 ).

Evaluate the function at these critical points and at the endpoints: ( f(0) = 0 ), ( f(2) = 2^3  12(2) = 16 ), ( f(4) = 4^3  12(4) = 16  48 = 32 ).

Compare the values obtained:
 ( f(0) = 0 )
 ( f(2) = 16 )
 ( f(4) = 32 )
The absolute maximum is ( f(0) = 0 ), and the absolute minimum is ( f(4) = 32 ).
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To locate the absolute extrema of the function ( f(x) = x^3  12x ) on the closed interval ([0,4]), you follow these steps:
 Find the critical points of the function within the interval by setting the derivative equal to zero and solving for (x).
 Evaluate the function at the critical points and the endpoints of the interval.
 Identify the highest and lowest values among these values to determine the absolute maximum and minimum.
Let's proceed with the steps:

Find the derivative of the function: [ f'(x) = 3x^2  12 ] Set ( f'(x) ) equal to zero and solve for ( x ) to find critical points: [ 3x^2  12 = 0 ] [ x^2  4 = 0 ] [ (x  2)(x + 2) = 0 ] [ x = 2 \text{ or } x = 2 ]

Evaluate the function ( f(x) ) at the critical points and the endpoints: [ f(0) = (0)^3  12(0) = 0 ] [ f(2) = (2)^3  12(2) = 16 ] [ f(4) = (4)^3  12(4) = 32 ]

Compare these values to find the absolute maximum and minimum: The values are: ( f(0) = 0 ), ( f(2) = 16 ), and ( f(4) = 32 ). The absolute maximum is ( f(0) = 0 ) and the absolute minimum is ( f(4) = 32 ).
So, the absolute maximum occurs at ( x = 0 ) and the absolute minimum occurs at ( x = 4 ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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