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

The equation of the tangent line is $y=1/4x+1$.
The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f(x)=\sqrt x\hspace{1cm}A(4,2)$$ 1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=4$ and $f(a)=b=2$ $$m=\lim_{h\to0}\frac{f(h+4)-2}{h}=\lim_{h\to0}\frac{\sqrt{h+4}-2}{h}$$ Multiply both numerator and denominator by $\sqrt{h+4}+2$: $$m=\lim_{h\to0}\frac{(\sqrt{h+4}-2)(\sqrt{h+4}+2)}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{h+4-4}{h(\sqrt{h+4}+2)}$$ $$m=\lim_{h\to0}\frac{h}{h(\sqrt{h+4}+2)}=\lim_{h\to0}\frac{1}{\sqrt{h+4}+2}$$ $$m=\frac{1}{\sqrt{0+4}+2}=\frac{1}{\sqrt4+2}=\frac{1}{4}$$ 2) Find the equation of the tangent line at $A(4,2)$: The tangent line would have this form: $$y=\frac{1}{4}x+m$$ Substitute $A(4,2)$ here to find $m$: $$\frac{1}{4}\times4+m=2$$ $$1+m=2$$ $$m=1$$ So the equation of the tangent line is $y=1/4x+1$.