# Is #f(x)=-x/e^(x^2-3x+2) # increasing or decreasing at #x=4 #?

And since the Product rule states

we can say

we can say

Then

Then

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To determine whether ( f(x) = -\frac{x}{e^{x^2 - 3x + 2}} ) is increasing or decreasing at ( x = 4 ), we need to examine the sign of its derivative at that point. We'll find the derivative of ( f(x) ) with respect to ( x ) using the quotient rule, then evaluate it at ( x = 4 ) and determine its sign.

[ f'(x) = \frac{-e^{x^2 - 3x + 2} - x \cdot e^{x^2 - 3x + 2} \cdot (2x - 3)}{(e^{x^2 - 3x + 2})^2} ]

Now, substituting ( x = 4 ) into ( f'(x) ) gives us:

[ f'(4) = \frac{-e^{4^2 - 3 \cdot 4 + 2} - 4 \cdot e^{4^2 - 3 \cdot 4 + 2} \cdot (2 \cdot 4 - 3)}{(e^{4^2 - 3 \cdot 4 + 2})^2} ]

[ f'(4) = \frac{-e^{16 - 12 + 2} - 4 \cdot e^{16 - 12 + 2} \cdot (8 - 3)}{(e^{16 - 12 + 2})^2} ]

[ f'(4) = \frac{-e^6 - 4 \cdot e^6 \cdot 5}{(e^6)^2} ]

[ f'(4) = \frac{-e^6 - 20e^6}{e^{12}} ]

[ f'(4) = \frac{-21e^6}{e^{12}} ]

Since ( e^6 ) and ( e^{12} ) are both positive, the sign of ( f'(4) ) depends only on the sign of ( -21 ). Since ( -21 ) is negative, ( f'(4) ) is negative.

Therefore, ( f(x) ) is decreasing at ( x = 4 ).

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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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