How do you find the global extreme values for #h(x)= x^23x# on [0,2]?
You find the extremes by taking the derivative and setting it to
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To find the global extreme values of ( h(x) = x^2  3x ) on the interval ([0, 2]), follow these steps:

Find the critical points of ( h(x) ) within the interval ([0, 2]) by taking the derivative and setting it equal to zero: [ h'(x) = 2x  3 ] [ 2x  3 = 0 ] [ x = \frac{3}{2} ]

Evaluate ( h(x) ) at the critical point and at the endpoints of the interval: [ h(0) = (0)^2  3(0) = 0 ] [ h(2) = (2)^2  3(2) = 2 ] [ h\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2  3\left(\frac{3}{2}\right) = \frac{9}{4}  \frac{9}{2} = \frac{9}{4} ]

Compare the values obtained in step 2. The maximum value is the largest value, and the minimum value is the smallest value.
Therefore, the global maximum value of ( h(x) ) on the interval ([0, 2]) is ( h(0) = 0 ), and the global minimum value is ( h\left(\frac{3}{2}\right) = \frac{9}{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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