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Question
find the slope of a line tangent to the curve y=3x^2 at the point P (1.5,6.75)by finding the limit of the slopes of the secant lines PQ where Q has x-values 1,1.4,1.49,1.499. can you show me step by step. thanks

Answer
We are given x1 = 1, x2 = 1.4, x3 = 1.49, and x4 = 1.499.
With these we get y1 = 3(1^2) = 2, y2 = 3(1.4^2) = 3*1.96 = 5.82,
y3 = 3(1.49^2) = 3*2.2201 = 6.6603, y4 = 3(1.499^2) = 3*2.247001 =
6.741003.

At x0 = 1.5, y0 = 3(1.5^2) = 3*2.25 = 6.75.

To approximate the slope, compute (yi-y0)/(xi-x0)
for i = 1, 2, 3, and 4.

For i = 1, we get (2-6.75)/(1 -1.5) = -4.75/-0.5 = 9.5.
These can be continued for i = 2, 3, and 4.

Again, the x's are 1, 1.4, 1.49, and 1.499.
The matching y's are 2, 5.82, 6.6603, and 6.741003.
For each one, x0 = 1.5 and y0 = 6.75.

Calculus

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