## Calculus: Early Transcendentals 8th Edition

(i) Using the first definition $m = \lim\limits_{x \to 1} = \frac{f(x) - f(1)}{x-1} = 2$ (ii) Using the second definition $m = \lim\limits_{h \to 0} = \frac{f(1+h) - f(1)}{h} = 2$ (b) Tangent line $y = 2x+1$
(i) Using the first definition $m = \lim\limits_{x \to 1} \frac{f(x) - f(1)}{x-1}$ $m = \lim\limits_{x \to 1} \frac{(4x-x^{2}) - (4(1)-1^{2})}{x-1}$ $m = \lim\limits_{x \to 1} \frac{-x^{2} +4x - 3}{x-1}$ $m = \lim\limits_{x \to 1} \frac{(x-1)(3-x)}{x-1}$ Cancel out $(x-1)$ $m = \lim\limits_{x \to 1} (3 -x)$ $m = (3 -1)$ $m = 2$ (ii) Using the second definition $m = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$ $m = \lim\limits_{h \to 0} \frac{[4(1+h) - (1+h)^{2}] - [4(1) - 1^{2}]}{h}$ $m = \lim\limits_{h \to 0} \frac{2h - h^{2}}{h}$ Simplify: $m = \lim\limits_{h \to 0} \frac{h(2- h)}{h}$ $m = \lim\limits_{h \to 0} 2 - h$ $m = \lim\limits_{h \to 0} 2 - 0$ $m = \lim\limits_{h \to 0} 2= 2$ (b) Tangent Line $m = \frac{y-3}{x-1}$ $m = 2$ $2 = \frac{y-3}{x-1}$ $2(x-1) = \frac{y-3}{x-1} (x-1)$ $2x - 2 = y-3$ $y = 2x -2+3$ $y = 2x + 1$