How to find instantaneous rate of change for #x^3 +2x^2 + x# at x from 1 to 2?
The instantaneous rate of change is the same as taking the derivative. This is defined at a point.
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To find the instantaneous rate of change of a function (f(x)) at a specific point (x = a), you can use the derivative (f'(a)). In this case, the function is (f(x) = x^3 + 2x^2 + x), and we want to find the instantaneous rate of change at (x = 1) and (x = 2).
To find the derivative of (f(x)), we first differentiate the function with respect to (x): [f'(x) = 3x^2 + 4x + 1]
Now, we can find the instantaneous rate of change at (x = 1) and (x = 2) by evaluating (f'(x)) at these points: [f'(1) = 3(1)^2 + 4(1) + 1 = 3 + 4 + 1 = 8] [f'(2) = 3(2)^2 + 4(2) + 1 = 3(4) + 8 + 1 = 12 + 8 + 1 = 21]
So, the instantaneous rate of change of (x^3 + 2x^2 + x) at (x = 1) is 8, and at (x = 2) is 21.
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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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