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THE SLOPE OF A TANGENT LINE
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[CHAP. 11
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Fig. 11-2 closer and closer to what we think of as the tangent line T to the graph at P. Hence, the slope of the line PQ will approach the slope of the tangent line at P; that is, the slope of the tangent line at P will be given by f (x + h) f (x) h 0 h lim What we have just said about tangent lines leads to the following precise de nition. De nition: Let a function f be continuous at x. By the tangent line to the graph of f at P(x, f (x)) is meant that line which passes through P and has slope
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f (x + h) f (x) h
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11.1 Consider the graph of the function f such that f (x) = x 2 (the parabola in Fig. 11-3). For a point P on the parabola having abscissa x, perform the calculations needed to nd f (x + h) f (x) h 0 h lim
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We have: f (x + h) = (x + h)2 = x 2 + 2xh + h2 f (x) = x 2 f (x + h) f (x) = (x 2 + 2xh + h2 ) x 2 = 2xh + h2 = h(2x + h) h(2x + h) f (x + h) f (x) = = 2x + h h h Thus,
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f (x + h) f (x) = lim (2x + h) = lim 2x + lim h = 2x + 0 = 2x h h 0 h 0 h 0
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and the slope of the tangent line at P is 2x. For example, at the point (2, 4), x = 2, and the slope of the tangent line is 2x = 2(2) = 4.
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CHAP. 11]
THE SLOPE OF A TANGENT LINE
11.2 Consider the graph of the function f such that f (x) = x 3 (see Fig. 11-4). For a point P on the graph having coordinates (x, x 3 ), compute the value of lim f (x + h) f (x) h
We have f (x + h) = (x + h)3 = x 3 + 3x 2 h + 3xh2 + h3 and f (x) = x 3 . algebra For any x and h, (x + h)3 = [(x + h)(x + h)](x + h) = (x 2 + 2xh + h2 )(x + h) = (x 3 + 2x 2 h + h2 x) + (x 2 h + 2xh2 + h3 ) = x 3 + 3x 2 h + 3xh2 + h3
Fig. 11-3
Fig. 11-4
Hence,
f (x + h) f (x) = (x 3 + 3x 2 h + 3xh2 + h3 ) x 3 = 3x 2 h + 3xh2 + h3 = h(3x 2 + 3xh + h2 ) h(3x 2 + 3xh + h2 ) f (x + h) f (x) = = 3x 2 + 3xh + h2 h h
f (x + h) f (x) = lim (3x 2 + 3xh + h2 ) h h 0 = 3x 2 + 3x(0) + 02 = 3x 2
This shows that the slope of the tangent line at P is 3x 2 . For example, the slope of the tangent line at (2, 8) is 3x 2 = 3(2)2 = 3(4) = 12.
11.3 (a) Find a formula for the slope of the tangent line at any point of the graph of the function f such that f (x) = 1/x (the hyperbola in Fig. 11-5). (b) Find the slope-intercept equation of the tangent line to the graph of f at the point (2, 1 ). 2
THE SLOPE OF A TANGENT LINE
[CHAP. 11
Fig. 11-5
1 x+h 1 x
f (x + h) =
f (x) =
f (x + h) f (x) =
1 1 x (x + h) x x h h = = = x+h x (x + h)x (x + h)x (x + h)x c ad bc ad bc a = = b d bd bd bd
algebra
Hence,
f (x + h) f (x) h h 1 1 = h= = h (x + h)x (x + h)x h (x + h)x lim 1 1 f (x + h) f (x) h 0 = lim = h lim (x + h)x h 0 h 0 (x + h)x lim
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