# What are the critical points of #h(x)=lnsqrt(3x-2(x^2))#?

So critical point is at

Set the derivative to zero to find critical points. To derive an ln function use this rule:

To derive u to find u' use the general power rule:

Now we have u', so plug it into the derivative of ln equation:

Which after some algebra will give you :

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Critical points exist when the derivative of the given point is 0 or undefined.

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To find the critical points of the function ( h(x) = \ln(\sqrt{3x - 2x^2}) ), we need to find where its derivative is equal to zero or undefined.

First, we find the derivative of ( h(x) ) using the chain rule and the derivative of the natural logarithm function:

[ h'(x) = \frac{1}{\sqrt{3x - 2x^2}} \cdot \frac{1}{2} \cdot (3 - 4x) ]

Setting ( h'(x) ) equal to zero and solving for ( x ), we get:

[ \frac{1}{\sqrt{3x - 2x^2}} \cdot \frac{1}{2} \cdot (3 - 4x) = 0 ]

This gives us two critical points: ( x = \frac{3}{4} ) and ( x = \frac{3}{2} ).

Now, we need to check for critical points where the derivative is undefined. The derivative is undefined when the denominator becomes zero. However, the expression ( \sqrt{3x - 2x^2} ) is always non-negative, so the derivative is never undefined.

Therefore, the critical points of ( h(x) ) are ( x = \frac{3}{4} ) and ( x = \frac{3}{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.

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