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

The tangent line whose slope is $1/4$ is $y=1/4x+1$.
$$y=f(x)=\sqrt{x}$$ The equation of the tangent line whose slope is $1/4$ has this form: $$y=\frac{1}{4}x+m$$ To find $m$, we need to locate the tangent point of this line, which is to find $x$ such that $$y'=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{1}{4}\hspace{1cm}(1)$$ First, we examine $f(x+h)-f(x)$: $$f(x+h)-f(x)=\sqrt{x+h}-\sqrt x$$ Therefore, $$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt x}{h}$$ Multiply both numerator and denominator by $\sqrt{x+h}+\sqrt x$: $$\frac{f(x+h)-f(x)}{h}=\frac{(\sqrt{x+h}-\sqrt x)(\sqrt{x+h}+\sqrt x)}{h(\sqrt{x+h}+\sqrt x)}=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt x)}$$ $$\frac{f(x+h)-f(x)}{h}=\frac{h}{h(\sqrt{x+h}+\sqrt x)}=\frac{1}{\sqrt{x+h}+\sqrt x}$$ Replacing the result into $(1)$: $$\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt x}=\frac{1}{4}$$ $$\frac{1}{\sqrt{x+0}+\sqrt x}=\frac{1}{4}$$ $$\frac{1}{2\sqrt x}=\frac{1}{4}$$ $$\sqrt x=2$$ $$x=4$$ For $x=4$, $y=f(4)=\sqrt4=2$ Substitute $x$ and $y$ for $m$: $$\frac{1}{4}\times4+m=2$$ $$1+m=2$$ $$m=1$$ The tangent line is $y=1/4x+1$.