Find the local linearization of g(x)=sqrt 4x near x=2
use local linearization to estimate g(2.05)
is your estimate too high or too low?
g(x) = sqrt 4x = 2 sqrtx
Local linearization near x = 2:
gl(x)= g(x=2) + [dg/dx@x=2])*(x-2)g
g(x=2) = 2 sqrt2
dg/dx = 2(1/2)/sqrtx = 1/sqrtx
dg/dx @ x=2 = 1/sqrt2
gl(x) = 2sqrt2 + (1/sqrt2)(x-2)
is the linearization about x = 2
For x = 2.05,
g(2.05) = 2.86356
The linearized approximation is
gl(2) + (0.05)/1.41421
= 2.8284 + 0.0353 = 2.86376
To find the local linearization of a function g(x) near a specific point x=a, you need to use the formula for the tangent line approximation, which is given by:
L(x) = g(a) + g'(a)(x-a)
Here, g'(a) represents the derivative of g(x) evaluated at x=a. In this case, g(x) = sqrt(4x) and the point of interest is x=2.
1. Calculate g'(x), the derivative of g(x). The derivative of sqrt(4x) can be found by applying the chain rule:
g'(x) = (1/2)(4x)^(-1/2) * 4
= 2 / sqrt(4x)
2. Evaluate g'(x) at x=2 to find g'(2):
g'(2) = 2 / sqrt(4*2)
= 2 / sqrt(8)
= 2 / (2 * sqrt(2))
= 1 / sqrt(2)
= sqrt(2) / 2
3. Plug these values into the formula for the local linearization to get L(x):
L(x) = g(2) + g'(2)(x-2)
Substituting g(2) = sqrt(4*2) = sqrt(8) = 2sqrt(2), the equation becomes:
L(x) = 2sqrt(2) + (sqrt(2)/2)(x-2)
4. Now, to estimate g(2.05) using local linearization, substitute x=2.05 into the equation for L(x):
L(2.05) = 2sqrt(2) + (sqrt(2)/2)(2.05-2)
= 2sqrt(2) + (sqrt(2)/2)(0.05)
= 2sqrt(2) + 0.025sqrt(2)
= (2+0.025)sqrt(2)
= 2.025sqrt(2)
Therefore, the estimate for g(2.05) using local linearization is approximately 2.025sqrt(2).
To determine if the estimate is too high or too low, you would need to compare this value to the actual value of g(2.05). However, since the actual function g(x) = sqrt(4x) is not provided, we cannot determine if the estimate is accurate.