Chapter 12, Problem 9PS

To determine

Find a function of the form $f(x)=a+b\sqrt{x}$ .

Answer to Problem 9PS

  $\boxed{f(x)=1+3\sqrt{x}}$

Explanation of Solution

Given information:

Find a function of the form $f(x)=a+b\sqrt{x}$ that is tangent to the line $2y-3x=5$ at the point $\left(1,4\right)$ .

Calculation:

Consider the following function and its tangent at the point $\left(1,4\right)$

  $f(x)=a+b\sqrt{x}$ ,

  $2y-3x=5$

To determine the values of the constants in the function, proceed as follows.

Differentiate the function.

  $\begin{array}{l}{f}^{\text{'}}(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(a+b\sqrt{x+h}\right)-\left(a+b\sqrt{x}\right)}{h}\end{array}$

  $=\underset{h\to 0}{\lim }\frac{b\left(\sqrt{x+h}-\sqrt{x}\right)}{h}$

Rationalize with $\left(\sqrt{x+h}+\sqrt{x}\right)$

  $f^{\text{'}}(x)=\underset{h\to 0}{\lim }\frac{b\left(\sqrt{x+h}-\sqrt{x}\right)}{h}\times \left[\frac{\left(\sqrt{x+h}+\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)}\right]$

  $\begin{array}{l}=\underset{h\to 0}{\lim }\frac{b\left(\left(x+h\right)-x\right)}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{h\to 0}{\lim }\frac{b\cancel{h}}{\cancel{h}\left(\sqrt{x+h}+\sqrt{x}\right)}\end{array}$

  $=\frac{b}{\sqrt{x}+\sqrt{x}}$

Hence, $\boxed{f^{\text{'}}(x)=\frac{b}{2\sqrt{x}}}$

Now find the value of the derivative at the point $\left(1,4\right)$

  $f^{\text{'}}(x)=\frac{b}{2\sqrt{x}}$

  $\begin{array}{l}m=f^{\text{'}}(1)\\ =\frac{b}{2}\end{array}$

Now find the equation of the tangent at the point $\left(1,4\right)$

  $\begin{array}{l}y-y_1=m(x-x_1)\\ y-4=\frac{b}{2}\left(x-1\right)\\ y=\frac{bx}{2}-\frac{b}{2}+4\end{array}$

Now rearrange the equation in the slope intercept form.

Slope-intercept form of a line $y=mx+c$

  $\begin{array}{l}2y-3x=5\\ 2y=3x+5\end{array}$

  $y=\frac{3}{2}x+\frac{5}{2}$

Now compare the equations.

  $\begin{array}{l}\frac{b}{2}=\frac{3}{2}\\ b=3\end{array}$

As the point $\left(1,4\right)$ lies on the graph of the function

  $\begin{array}{l}f(x)=a+b\sqrt{x}\\ f\left(1\right)=a+b\sqrt{1}\\ 4=a+b\end{array}$

Now substitute the value $b$ and solve for a

  $\begin{array}{l}4=a+3\\ a=1\end{array}$

Hence, the expression for the function is $\boxed{f(x)=1+3\sqrt{x}}$ .