What is the instantaneous rate of change of #f(x)=x/(-x-8)# at #x=4 #?
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To find the instantaneous rate of change of (f(x) = \frac{x}{-x-8}) at (x = 4), we'll first find the derivative of the function (f(x)) with respect to (x), and then evaluate it at (x = 4).
Given (f(x) = \frac{x}{-x-8}), we'll use the quotient rule to find its derivative:
[f'(x) = \frac{d}{dx}\left(\frac{x}{-x-8}\right)]
Using the quotient rule:
[f'(x) = \frac{(1)(-x - 8) - (x)(-1)}{(-x - 8)^2}]
[f'(x) = \frac{-x - 8 + x}{(-x - 8)^2}]
[f'(x) = \frac{-8}{(-x - 8)^2}]
Now, we'll evaluate the derivative (f'(x)) at (x = 4):
[f'(4) = \frac{-8}{(-4 - 8)^2}]
[f'(4) = \frac{-8}{(-12)^2}]
[f'(4) = \frac{-8}{144}]
[f'(4) = -\frac{1}{18}]
So, the instantaneous rate of change of (f(x)) at (x = 4) is (-\frac{1}{18}).
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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.
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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