How do you use the limit definition to find the slope of the tangent line to the graph #y = x^(3) + 2x^(2) - 3x + 2# at x=1?
See below.
but I don't really want to get into cubing a binomial. So I'll use
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To find the slope of the tangent line to the graph y = x^3 + 2x^2 - 3x + 2 at x = 1 using the limit definition, we can follow these steps:
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Start with the given function: y = x^3 + 2x^2 - 3x + 2.
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Calculate the difference quotient, which represents the slope of the secant line between two points on the graph. Let's choose a point close to x = 1, such as x = 1 + h, where h is a small positive number.
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Substitute the chosen point into the function to find the corresponding y-value: y = (1 + h)^3 + 2(1 + h)^2 - 3(1 + h) + 2.
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Expand and simplify the expression obtained in step 3.
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Subtract the original function evaluated at x = 1 from the expression obtained in step 4.
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Divide the result from step 5 by h.
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Take the limit as h approaches 0 of the expression obtained in step 6.
The limit obtained in step 7 will give us the slope of the tangent line to the graph at x = 1.
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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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- How do you find the tangent line of #f(x) = 3-2x # at x=-1?

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