Chapter 12, Problem 3PS

To determine

The function of the form $f\left(x\right)=a+b\sqrt{x}$ that is tangent to the given line at given point.

Answer to Problem 3PS

The $f\left(x\right)=1+3\sqrt{x}$

Explanation of Solution

Given:

The given line is,

  $2y-x-5$

The given point is,

  $\left(1,4\right)$

Concept used:

The slope of the tangent line to a function is given by,

  $m=f^{\prime }\left(x\right)\dots \left(1\right)$

The slope intercept form of a line is given by,

  $y=mx+c\dots \left(2\right)$

Calculation:

Write the given line in slope intercept form.

  $\begin{array}{l}2y=3x+5\\ y=\frac{3}{2}x+\frac{5}{2}\\ \text{Comare}\text{with}\text{equation}\left(\text{2}\right)\\ m=\frac{3}{2}\end{array}$

Differentiate the given function.

  $\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{b}{2\sqrt{x}}\\ \text{Substitute}\text{the}\text{value,}\\ f^{\prime }\left(1\right)=\frac{b}{2\sqrt{1}}\\ =\frac{b}{2}\end{array}$

From equation (1),

  $\begin{array}{l}m{=}^{\prime }f\left(x\right)\\ \text{Substitute}\text{values,}\\ \frac{3}{2}=\frac{b}{2}\\ b=3\end{array}$

The given point lies on the function so it will satisfy its equation.

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

Substitute the value in given function.

  $f\left(x\right)=1+3\sqrt{x}$

Conclusion:

Hence, the function is $f\left(x\right)=1+3\sqrt{x}$ .