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

x y = f(x) m(x) = the estimated SLOPE of the tangent line to y=f(x) at the point (x,y)
0 2 –1 
1 1 –1 
2 1/3 0
3 1 1
4 3/2 1/2
5 1 –2


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.