Chapter 5.5, Problem 10QR

To determine

To graph:

  $y=f\left(x\right)$ whose tangent line at $x=c$

Explanation of Solution

Given information:

  $f\left(x\right)=\left\{\begin{array}{l}-\sqrt{3-x}\text{ , }x<3\\ \sqrt{x-3}\text{ , }x\ge 3\end{array}\right\};c=4$

Graph:

To find the equation of the tangent at $x=c=4$

  $\begin{array}{l}f\left(x\right)=\sqrt{x-3}\\ f\text{'}\left(x\right)=\frac{1}{2}{\left(x-3\right)}^{-\frac{1}{2}}=\frac{1}{2\sqrt{\left(x-3\right)}}\end{array}$

The slope of the tangent is at $x=4$ so,

  $f\text{'}\left(4\right)=\frac{1}{2\sqrt{\left(4-3\right)}}=\frac{1}{2}$

Therefore, equation of the tangent to curve $f\left(x\right)$ at $x=c=4$ will be,

  $\begin{array}{l}y=f\left(4\right)+\frac{1}{2}\left(x-4\right)\\ y=\sqrt{4-3}+\frac{1}{2}\left(x-4\right)\\ y=1+\frac{x}{2}-2\\ y=\frac{x}{2}-1\end{array}$

  

Interpretation:

The graph is obtained by using the data which calculated from the given information. The obtained data is plotted in the graph