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Question 968481: Please help,
Consider the function f(x) = x2 − 1. Find the equation of the tangent to the graph of f(x) at x = 3. [NOTE: when calculating f′(3), use first principles.]
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! I assume the function f(x) = x2 − 1 is really
f(x) = x^2 - 1
we need the first derivative of this function to find the slope
using first principles,
f'(x) = lim h->0 (f(x+h) - f(x))/h
now let us find f(x+h)
f(x+h) = (x+h)^2 - 1 = x^2 + 2xh + h^2 - 1
f(x+h) - f(x) = x^2 + 2xh + h^2 - 1 -x^2 + 1 = 2xh + h^2
(f(x+h) - f(x))/h = 2x + h
as h->0 we can drop the h and limit becomes 2x
so f'(x) = 2x
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therefore slope at x = 3 is
f'(3) = 2*3 = 6
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using f(x) = x^2 - 1 to find y when x = 3
y = 3^2 -1 = 8
therefore our point is (3, 8), so far we have
y = 6x + b
now find b when x = 3 and y = 8
8 = 6*3 + b
b = -10
and the equation of the line tangent to the function at x = 3 is
y = 6x - 10
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