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Extra info for Instructor's Solutions Manuals for Calculus Early Transcendentals 7th Edition
005: Because f (x) = 0 · x2 + 4x − 5, the slope-predictor for f is m(a) = 2 · 0 · a + 4 = 4. So the line tangent to the graph of f at (2, f (2)) has slope 4 and therefore equation y − 3 = 4(x − 2); that is, y = 4x − 5. 006: Because f (x) = 0 · x2 − 3x + 7, the slope-predictor for f is m(a) = 2 · 0 · a − 3 = −3. So the line tangent to the graph of f at (2, f (2)) has slope −3 and therefore equation y − 1 = −3(x − 2); that is, y = −3x + 7. 007: Because f (x) = 2x2 − 3x + 4, the slope-predictor for f is m(a) = 2 · 2 · a − 3 = 4a − 3.
A line tangent to the graph of f will be horizontal when m(a) = 0, thus when a = 10. So the tangent line is horizontal at the point (10, 0) and at no other point of the graph of f . 025: If f (x) = x2 , then the slope-predictor for f is m(a) = 2a. So the line tangent to the graph of f at the point P (−2, 4) has slope m(−2) = −4 and the normal line at P has slope 14 . Hence an equation for the line tangent to the graph of f at P is y − 4 = −4(x + 2); that is, y = −4x − 4. An equation for the line normal to the graph of f at P is y − 4 = 14 (x + 2); that is, 4y = x + 18.
1. 10−4 10−6 10−8 The limit appears to be 1. 499817 0 Beware of round-oﬀ errors. 5. 166667 The limit appears to be 16 . 00001 1. 1. 999986 1. 1. The limit appears to be 1. 71828 (1 + x)1/x 4. 01] is next. 059: lim x→0 x − tan x 1 =− . 3333. sin 2x 2 = . 060: lim π π = 0 for every positive integer n. Therefore lim sin , if it x→0 x π were to exist, would be 0. Notice however that sin 3n · alternates between +1 and −1 for n = 1, 2, 3, . . 2 π Therefore lim sin does not exist. 00001] is shown next.