Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 95

Solve for the value of constant $c$ such that the line $\displaystyle y = \frac{3}{2}x + 6$ is tangent to the
curve $y = c \sqrt{x}$

If the line is tangent to the curve at some point, it means that the first derivative of the curve is equal
to the slope of the line and by inspection, the slope of the line $\displaystyle \frac{3}{2}$ by the formula
$y = mx + b $, where $m$ is the slope.

Taking the first derivative of the curve with respect to $x$

$
\begin{equation}
\begin{aligned}
y &= c \sqrt{x}\\
\\
y'&= \text{slope} = c \left( \frac{1}{2\sqrt{x}}\right)\\
\\
\frac{c}{\cancel{2}\sqrt{x}} &= \frac{3}{\cancel{2}} && \text{Using cross multiplication}\\
\\
c &= 3\sqrt{x}
\end{aligned}
\end{equation}
$


Also, if the line is tangent to the curve, they instersect at one point and the line passes through
the curve therefore we can equate both equations.
$\displaystyle c \sqrt{x} = \frac{3}{2}x + 6$
but the value of $c = 3\sqrt{x}$
$ \displaystyle 3 \sqrt{x}(\sqrt{x}) = \frac{3}{2}x +6$
$\displaystyle 3x = \frac{3}{2}x +6 $

Solving for $x$:
$x = 4$

Now, solving for $c$, we substitute the value of $x$

$
c = 3 \sqrt{x}\\
c = 3 \sqrt{4}\\
\boxed{c = 6}
$