15. -/1 points details my notes find the value of c such that the line y = 7/6x + 21/2 is tangent to the…

Explanation:

Step1: Differentiate the curve

The derivative of $y = c\sqrt{x}=cx^{\frac{1}{2}}$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=\frac{c}{2\sqrt{x}}$.

Step2: Equate the slope of the tangent line

The slope of the line $y=\frac{7}{6}x+\frac{21}{2}$ is $\frac{7}{6}$. Set $y'=\frac{7}{6}$, so $\frac{c}{2\sqrt{x}}=\frac{7}{6}$, which gives $c=\frac{7\sqrt{x}}{3}$.

Step3: Find the intersection point

Since the line $y=\frac{7}{6}x+\frac{21}{2}$ and the curve $y = c\sqrt{x}$ are tangent, at the point of tangency their $y$ - values are equal. Substitute $c=\frac{7\sqrt{x}}{3}$ into $y = c\sqrt{x}$, we get $y=\frac{7\sqrt{x}}{3}\cdot\sqrt{x}=\frac{7x}{3}$. Also, $y=\frac{7}{6}x+\frac{21}{2}$. Then $\frac{7x}{3}=\frac{7}{6}x+\frac{21}{2}$.

Step4: Solve for $x$

Multiply through by 6 to clear the fractions: $14x = 7x+63$. Subtract $7x$ from both sides: $14x - 7x=63$, so $7x = 63$ and $x = 9$.

Step5: Solve for $c$

Substitute $x = 9$ into $c=\frac{7\sqrt{x}}{3}$. Since $\sqrt{9}=3$, then $c=\frac{7\times3}{3}=7$.

Answer:

$7$