Download Section 2.8: Linear Approximation An Example

Section 2.8: Linear Approximation
An Example
Given: a function and an x‐value we’ll call The Linearization function is Note: the linearization function is really just the line tangent to is called the linear approximation of at the point because (1) It is a linear function. (ie: if we rearrange the equation above, ) (2) For x‐values which are close to , . (ie: to a certain number of decimal places.) Example: given the and Find the linearization of 2 at √
2. Then use it to approximate the numbers √3.98 and √4.05. Lastly, for what values of x is the linear approximation accurate to within 0.5? (a) Find : 2
⟹ 1
∙
2
/
2
2
1
4
1
∙ 1
2
⟹ /
1
4
2√
2
2
2 2
1
4
3
2
(b) Find √3.98 and √4.05: We want to use the fact that √3.98
1
4
√2
⟹ √3.98
for x‐values near 2 1.98
1
1.98
4
3
⟹ 2
3
2
1.995 The actual answer, if you plug it into a calculator is √3.98
√4.05
1
4
√2
⟹ √4.05
2.05
1
2.05
4
1.98 1.99499 3
⟹ 2
3
2
2.05 2.0125 The actual answer, if you plug it into a calculator is √4.05
2.01246 √3.9 √3.98 √4 √4.05 √4.1 √5 √6 2? √
(c) How accurate is 1.9 1.98 2 2.05 2.1 1 2 Actual value 1.974841766 1.994993734 2 2.01246118 2.024845673 2.236067977 2.449489743 1.975 1.995 2 2.0125 2.025 2.25 2.5 Accurate to within? 0.001 0.0001 0 0.0001 0.001 0.015 0.015 (d) Accurate to within 0.5: If we want to be accurate to within 0.5 of |
|
0.5 0.5
0.5
0.5 1
1
4
2
√
0.5
, this means that we want: 2
√
2
√
√
If you graph the function 3
2
4
4
2
3
2
0.5 0.5 2 , you’ll find that its y‐values are between 1 and 2 when the x‐values are between: 1.657
9.65685 Result: for this problem, any time you use L(x) to approximate f(x) you’ll be within 0.5 of the actual answer for any x you have that is in the above interval.