## Calculus: Early Transcendentals 8th Edition

$c=6$ is the result of this exercise.
$$(l): y=\frac{3}{2}x+6$$ The slope of line $(l)$ is $\frac{3}{2}$. $$(S): y=f(x)=c\sqrt x$$ $$(S): y=f(x)=cx^{1/2}$$ - Derivative of $f(x)$: $$f'(x)=\frac{1}{2}cx^{-1/2}$$ $$f'(x)=\frac{c}{2\sqrt x}$$ Line $(l)$ can only be tangent with curve $(S)$ if both of the followings happen: 1) Line $(l)$ intersects curve $(S)$ at at least one point $A(a, b)$. 2) At point A, $f'(a)$ equals the slope of line $(l)$. *Consider 1): If $A(a,b)$ is where line $(l)$ and curve $(S)$ intersects, then $A$ must lie in both the line and the curve. Therefore, $$b=c\sqrt a\hspace{1cm}(1)$$ and $$b=\frac{3}{2}a+6\hspace{1cm}(2)$$ Consider 2): At point $A(a,b)$:$$f'(a)=\frac{3}{2}$$ $$\frac{c}{2\sqrt a}=\frac{3}{2}$$ $$\frac{c}{\sqrt a}=3$$ $$c=3\sqrt a\hspace{1cm} (3)$$ Substitute (3) into (1), we have $$b=3\sqrt a\sqrt a=3(\sqrt a)^2=3a\hspace{1cm}(4)$$ Substitue (4) into (2), we have $$3a=\frac{3}{2}a+6$$ $$\frac{3}{2}a=6$$ $$a=4$$ Substitute $a=4$ to (3), we have $$c=3\sqrt4=6$$ $c=6$ is the result.