MA005: Calculus I

Practice 1: Approximate values of $m(x)$ are in the table below. Fig. 17 is a graph of $m(x)$.

Practice 2: The tangent lines to the graph of g are horizontal $(slope = 0)$ when $x ≈ –1, 1, 2.5$, and $5$. 

Practice 3: Fig. 18 is a graph of the approximate rate of temperature change (slope). 

Practice 4: $\mathrm{y}=\mathrm{x}^{2}$

$\begin{aligned} &\operatorname{At}(1,1), \mathrm{m}_{\tan }=\lim\limits_{h \rightarrow 0} \frac{f(1+h)-f(1)}{(1+h)-(1)}=\lim\limits_{h \rightarrow 0} \frac{(1+h)^{2}-(1)^{2}}{h}=\lim\limits_{h \rightarrow 0} \frac{\left\{1+2 h+h^{2}\right\}-1}{h} \\ &=\lim\limits_{h \rightarrow 0} \frac{2 h+h^{2}}{h}=\lim\limits_{h \rightarrow 0} \frac{h(1+h)}{h}=\lim\limits_{h \rightarrow 0}(2+h)=2 \end{aligned}$

$\mathrm{At}(0,0), \mathrm{m}_{\tan }=\lim\limits_{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-(0)}=\lim\limits_{h \rightarrow 0} \frac{(0+h)^{2}-(0)^{2}}{h}=\lim\limits_{h \rightarrow 0} \frac{h^{2}}{h}=\lim\limits_{h \rightarrow 0} \mathrm{~h}=\mathbf{0}$

$\operatorname{At}(-1,1), \mathrm{m}_{\tan }=\lim\limits_{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{(-1+h)-(-1)}=\lim\limits_{h \rightarrow 0} \frac{\left\{1-2 h+h^{2}\right\}-1}{h}=\lim\limits_{h \rightarrow 0} \frac{-2 h+h^{2}}{h}=-\mathbf{2}$

Practice 5: From Example 4 we know the slope of the tangent line is $m_{\tan }=2 x$ so the slope of the tangent line at $(2,4)$ is $m_{\tan }=2 x=2(2)=4$. The tangent line has slope 4 and goes through the point $(2,4)$ so the equation of the tangent line (using $\left.y-y_{o}=m\left(x-x_{o}\right)\right)$ is $y-4=4(x-2)$ or $y=4 x-4$. The point $(3,8)$ satisfies the equation $\mathrm{y}=4 \mathrm{x}-4$ so the point ((3,8)\) lies on the tangent line.