# Thread: finding corresponding values of L9(x)

1. ## finding corresponding values of L9(x)

http://www.flickr.com/photos/4882931...18204/sizes/l/

Its number 5.
The value was changed to plus/minus 0.05
How should i go about solving this problem?

2. bump...
anyone? i need to have this figured out by tomorrow >.<

3. Since no one here has any idea what you mean by "L9(x)", I don't see how anyone can help you.

4. It means the linearization (tangent line) at x=9

5. The derivative of $\displaystyle \sqrt{x}= x^{1/2}$ is $\displaystyle (1/2)x^{-1/2}= \frac{1}{2\sqrt{x}}$. At x= 9 that is $\displaystyle \frac{1}{2\sqrt{9}}= \frac{1}{6}$ while [tex]\sqrt{9}= 3[/math so the tangent line at x= 9 is $\displaystyle y= \frac{1}{6}(x- 3)+ 3$. In order that that be within $\displaystyle \pm .05$ of the actual value, we must have $\displaystyle |\sqrt{x}- \frac{1}{6}(x- 3)+ 3|\le .05$ or $\displaystyle -.05\le \sqrt{x}- \frac{1}{6}(x- 3)+ 3\le .05$. Solve $\displaystyle \sqrt{x}- \frac{1}{6}(x- 3)+ 3= -.05$ and $\displaystyle sqrt{x}- \frac{1}{6}(x- 3)+ 3= .05$.