What are the local extrema, if any, of #f (x) = x^3-12x+2 #?
The function has 2 extrema:
To find extrema we calculate derivative
Now we have to check if the derivative changes sign at the calcolated points:
graph{x^2-4 [-10, 10, -4.96, 13.06]}
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To find the local extrema of ( f(x) = x^3 - 12x + 2 ), we first need to find the critical points by setting the derivative equal to zero and solving for ( x ). Then we can determine if these critical points correspond to local maxima or minima.
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Find the derivative of ( f(x) ): [ f'(x) = 3x^2 - 12 ]
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Set ( f'(x) ) equal to zero and solve for ( x ): [ 3x^2 - 12 = 0 ] [ x^2 - 4 = 0 ] [ x = \pm 2 ]
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Determine the nature of the critical points:
- To the left of ( x = -2 ), ( f'(x) ) changes from negative to positive, indicating a local minimum.
- To the right of ( x = -2 ), ( f'(x) ) changes from positive to negative, indicating a local maximum.
- To the left of ( x = 2 ), ( f'(x) ) changes from negative to positive, indicating a local minimum.
- To the right of ( x = 2 ), ( f'(x) ) changes from positive to negative, indicating a local maximum.
Therefore, the local extrema of ( f(x) ) are:
- Local minimum at ( x = -2 )
- Local maximum at ( x = 2 )
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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.
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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