# How do you find the critical points for #f(x) = x^3 - 15x^2 + 4# and the local max and min?

The critical points are:

To identify the critical points, we have to solve the equation:

that is:

graph{x^3-15x^2+4 [-80, 80, -640, 640]}

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To find the critical points of ( f(x) = x^3 - 15x^2 + 4 ), we first find its derivative, then set it equal to zero to solve for critical points.

The derivative of ( f(x) ) is ( f'(x) = 3x^2 - 30x ).

Setting ( f'(x) ) equal to zero, we have ( 3x^2 - 30x = 0 ).

Factoring out ( 3x ) gives ( 3x(x - 10) = 0 ).

This yields critical points ( x = 0 ) and ( x = 10 ).

To determine whether these critical points correspond to local maxima, minima, or neither, we can use the second derivative test or analyze the behavior of the function around these points.

Taking the second derivative of ( f(x) ) yields ( f''(x) = 6x - 30 ).

Evaluating ( f''(0) ) and ( f''(10) ):

At ( x = 0 ), ( f''(0) = -30 ), indicating a concave down shape, implying a local maximum.

At ( x = 10 ), ( f''(10) = 30 ), indicating a concave up shape, implying a local minimum.

Therefore, ( f(x) = x^3 - 15x^2 + 4 ) has a local maximum at ( x = 0 ) and a local minimum at ( x = 10 ).

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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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