# Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 119: 50 $$f(x)=x+\frac{5}{x}\hspace{1cm}x_0=1$$ a) Here, since $x_0=1$, we will plot $y=f(x)$ over the interval $1/2\le x\le4$. The graph of $y=f(x)$ is represented by the red curve. b) and c) With $x_0=1$: $$q(h)=\frac{f(h+1)-f(1)}{h}=\frac{f(h+1)-(1+\frac{5}{1})}{h}=\frac{f(h+1)-6}{h}$$ The graph of $q(h)$ is also plotted and represented by the purple line. As $h\to0$, from the graph, we see that $q$ approaches $-4$. Therefore, $$\lim_{h\to0}q(h)=-4$$ d) For $x_0=1$: $$y=f(1)+q(h)\times(x-1)=6+q(h)\times(x-1)$$ - With $h=3$: $$y=6+q(3)\times(x-1)=6+\frac{f(4)-6}{3}(x-1)=6+\frac{(4+\frac{5}{4}-6)(x-1)}{3}$$ $$y=6-\frac{1}{4}(x-1)=6-\frac{1}{4}x+\frac{1}{4}=-\frac{1}{4}x+\frac{25}{4}$$ This secant line is represented by the black line. - With $h=2$: $$y=6+q(2)\times(x-1)=6+\frac{f(3)-6}{2}(x-1)=6+\frac{(3+\frac{5}{3}-6)(x-1)}{2}$$ $$y=6-\frac{2}{3}(x-1)=6-\frac{2}{3}x+\frac{2}{3}=-\frac{2}{3}x+\frac{20}{3}$$ This secant line is represented by the blue line. - With $h=1$: $$y=6+q(1)\times(x-1)=6+\frac{f(2)-6}{1}(x-1)=6+\Big(2+\frac{5}{2}-6\Big)(x-1)$$ $$y=6-\frac{3}{2}(x-1)=6-\frac{3}{2}x+\frac{3}{2}=-\frac{3}{2}x+\frac{15}{2}$$ This secant line is represented by the green line. - The tangent line at $x_0=1$: The tangent line at $x=0$ will have this form: $$y=\lim_{h\to0}q(h)x+m=-4x+m$$ To find m, we use the fact that the tangent line is at $x=1$: $$-4\times1+m=f(1)=6$$ $$-4+m=6$$ $$m=10$$ So the tangent line at $x_0$ is $y=-4x+10$. It is represented by the orange line. 