# Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises - Page 113: 4

a.) i.) $f'(1)=-2$ $\quad$ ii.) $f'(1)=-2$ b.) $y=-2x+2$ c.) See attached file

#### Work Step by Step

a.) i.) $\lim\limits_{x \to 1} \dfrac{f(x) - f(1)}{x - 1} = \lim\limits_{x \to 1} \dfrac{x - x^3 - (1 - (1)^3)}{x - 1} = \lim\limits_{x \to 1} \dfrac{x - x^3 - (1 - 1)}{x - 1} = \lim\limits_{x \to 1} \dfrac{x - x^3 - 0}{x - 1} = \lim\limits_{x \to 1} \dfrac{x - x^3}{x - 1} = \lim\limits_{x \to 1} \dfrac{-x(-1+x^2)}{x - 1} = \lim\limits_{x \to 1} \dfrac{-x(x^2-1)}{x - 1} = \lim\limits_{x \to 1} \dfrac{-x(x+1)(x-1)}{x - 1} =\\ \lim\limits_{x \to 1} -x(x+1) = -1(1+1) = -(2) = -2 \longrightarrow f'(1)=-2$ ii.) $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h} = \lim\limits_{h \to 0} \dfrac{(x+h)-(x+h)^3-(x-x^3)}{h} = \lim\limits_{h \to 0} \dfrac{x+h-(x^3+3x^2h+3xh^2+h^3)-x+x^3}{h} = \lim\limits_{h \to 0} \dfrac{h-x^3-3x^2h-3xh^2-h^3+x^3}{h} = \lim\limits_{h \to 0} \dfrac{h-3x^2h-3xh^2-h^3}{h} = \lim\limits_{h \to 0}\dfrac{h(1-3x^2-3xh-h^2)}{h} = \lim\limits_{h \to 0}(1-3x^2-3xh-h^2) = 1-3x^2-3x0-(0)^2 =\\ 1-3x^2-0-0 = 1-3x^2 \longrightarrow f'(x) = 1-3x^2$ Therefore the slope of the tangent line to the curve $y=x-x^3$ at the point $(1,0)$ is: \begin{gather*} f'(1) = 1-3(1)^2 = 1-3 = -2 \end{gather*} b.)Due to the point-slope equation, the tangent line to the curve $y=f(x)$ that goes through the point $(a,f(a))$ and has slope $f'(a)$ is: \begin{gather*} y-f(a) = f'(a)(x-a) \end{gather*}Replacing, we have: \begin{gather*} y-0=-2(x-1) \longrightarrow y=-2x+2 \end{gather*}

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