finding corresponding values of L9(x)

**prettyfunny31** · March 30th 2010, 09:48 AM

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?

**prettyfunny31** · March 30th 2010, 06:41 PM

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

**HallsofIvy** · March 31st 2010, 02:00 AM

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

**prettyfunny31** · March 31st 2010, 05:18 AM

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

**HallsofIvy** · March 31st 2010, 12:49 PM

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

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