## University Calculus: Early Transcendentals (3rd Edition)

There is a tangent line which satisfies the request of the exercise, whose equation is $$y=\frac{1}{2}x+\frac{1}{2}$$ and the point of tangency is $(1,1)$.
$$y=f(x)=\sqrt x$$ First, we will find the derivative $y'$: $$y'=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{\sqrt z-\sqrt x}{z-x}$$ Multiply both numerator and denominator by $\sqrt z+\sqrt x$: $$y'=\lim_{z\to x}\frac{(\sqrt z-\sqrt x)(\sqrt z+\sqrt x)}{z-x(\sqrt z+\sqrt x)}=\lim_{z\to x}\frac{z-x}{z-x(\sqrt z+\sqrt x)}$$ $$y'=\lim_{z\to x}\frac{1}{\sqrt z+\sqrt x}$$ $$y'=\frac{1}{\sqrt x+\sqrt x}=\frac{1}{2\sqrt x}$$ Here we need to find a tangent line to $f(x)$ which also crosses the $x$-axis at $x=-1$, or in fact, crosses the point $A(-1,0)$. Let's call this tangent $(t)$. $(t)$ also crosses the point of tangency, which, because it lies in the graph of $f(x)$ as well, arbitrarily $B(x,\sqrt x)$. Using these 2 points $A$ and $B$, we can have a formula for the slope of $(t)$: $$s=\frac{\sqrt x-0}{x-(-1)}=\frac{\sqrt x}{x+1}$$ But $s=y'=\frac{1}{2\sqrt x}$ as well. Therefore, $$\frac{\sqrt x}{x+1}=\frac{1}{2\sqrt x}$$ $$\frac{2x-(x+1)}{2\sqrt x(x+1)}=0$$ $$x-1=0$$ $$x=1$$ So the slope $s$ is $s=\frac{1}{2\sqrt1}=\frac{1}{2}$ and the point of tangency is $B(1,1)$. The equation of our tangent line $(t)$ has this form: $$y=\frac{1}{2}x+m$$ To find $m$, we rely on the fact that $(t)$ passes $A(-1,0)$: $$\frac{1}{2}\times(-1)+m=0$$ $$-\frac{1}{2}+m=0$$ $$m=\frac{1}{2}$$ So the equation of $(t)$ is $$y=\frac{1}{2}x+\frac{1}{2}$$