# What are the local extrema, if any, of #f (x) =(x^3+3x^2)/(x^2-5x)#?

By the Quotient rule we get

so we have

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To find the local extrema of ( f(x) = \frac{x^3 + 3x^2}{x^2 - 5x} ), we first need to find the critical points by taking the derivative and setting it equal to zero. After that, we will determine whether these critical points correspond to local extrema.

The derivative of ( f(x) ) is ( f'(x) = \frac{(3x^2 + 6x)(x^2 - 5x) - (x^3 + 3x^2)(2x - 5)}{(x^2 - 5x)^2} ).

Now, setting ( f'(x) ) equal to zero and solving for ( x ) gives us the critical points.

After solving for ( x ), we obtain ( x = 0 ) and ( x = -3 ) as the critical points.

Next, we evaluate the second derivative ( f''(x) ) to determine the nature of these critical points.

The second derivative of ( f(x) ) is ( f''(x) = \frac{24x^2 - 120x}{(x^2 - 5x)^2} ).

Now, we evaluate ( f''(0) ) and ( f''(-3) ) to determine the concavity at these critical points.

( f''(0) = 0 ), so ( x = 0 ) is an inflection point rather than a local extremum.

( f''(-3) = 72 ), which is positive, indicating that ( x = -3 ) corresponds to a local minimum.

Therefore, the function ( f(x) = \frac{x^3 + 3x^2}{x^2 - 5x} ) has a local minimum at ( x = -3 ), and there are no other local extrema.

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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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