# linearization

• Oct 28th 2009, 05:33 PM
hazecraze
linearization
Find the linear approximation of the function f(x) = √(25-x) at a = 0 and use it to approximate the numbers √24.9 and √24.99.

L(x)=

$L(x)=f(a)+f'(a)(x-a)$

$f'(a)=\frac{1}{2sqrt(25-x)}$
$f'(0)=\frac{1}{10}$
$f(a)=5$
$
L(x)=5 +\frac{1}{10}(x-25)$

(I used this to get the
√24.9 and √24.99; (4.99, 4.999 respectively), but...

$
L(x)=\frac{50}{10} + \frac{1}{10}x - \frac{25}{10}$

$L(x)=$ http://www.webassign.net/cgi-bin/pal...E%3C%2Fmath%3E

I put this in for part a, and it was marked wrong. A=0, but if I put (x-0) instead of (x-25) it becomes a number greater than 5, which won't make sense considering you plug in a number lower than 25.
• Oct 28th 2009, 05:43 PM
skeeter
Quote:

Originally Posted by hazecraze
Find the linear approximation of the function f(x) = √(25-x) at a = 0 and use it to approximate the numbers √24.9 and √24.99.

L(x)=

$L(x)=f(a)+f'(a)(x-a)$

$f'(a)=\frac{1}{2sqrt(25-x)}$ ***
$f'(0)=\frac{1}{10}$
$f(a)=5$
$
L(x)=5 +\frac{1}{10}(x-25)$

(I used this to get the
√24.9 and √24.99; (4.99, 4.999 respectively), but...

$
L(x)=\frac{50}{10} + \frac{1}{10}x - \frac{25}{10}$

$L(x)=$ http://www.webassign.net/cgi-bin/pal...E%3C%2Fmath%3E

I put this in for part a, and it was marked wrong. A=0, but if I put (x-0) instead of (x-25) it becomes a number greater than 5, which won't make sense considering you plug in a number lower than 25.

noticed a sign error ***

$f(x) = \sqrt{25-x}$

$f'(x) = -\frac{1}{2\sqrt{25-x}}$
• Oct 29th 2009, 10:44 AM
hazecraze
I set it up in point slope form like you would for the tangent line and got:
$(y-5)=(\frac{-1}{10}(x-0))$
$y= 5 + \frac{-1}{10}x$

So x if really delta(x)? (25-24.9 or 24.99)
• Oct 29th 2009, 04:30 PM
skeeter
$f(x) = \sqrt{25-x}$

$f'(x) = -\frac{1}{\sqrt{25-x}}$

$f(.1) = \sqrt{24.9}$

$f(.1) \approx L(.1) = f(0) + f'(0)(.1 - 0) = 5 - \frac{1}{5} \cdot (0.1) = 4.98$

$f(.01) = \sqrt{24.99}$

$f(.01) \approx L(.01) = f(0) + f'(0)(.01 - 0) = 5 - \frac{1}{5} \cdot (0.01) = 4.998$