Verify the given linear approximation at a = 0.Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.)

(1+2x)^(1/4)¡Ö 1 +(1/2)x

To verify the given linear approximation at a = 0, we will calculate the linear approximation using the formula:

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

Where f(x) is the original function, f'(x) is the derivative of f(x), a is the point of approximation, and x is the value for which we want to find the linear approximation.

First, let's find the derivative of the original function f(x) = (1+2x)^(1/4):

f'(x) = (1/4)(1+2x)^(-3/4)(2)

= (1/4)(2)(1+2x)^(-3/4)

= (1/2)(1+2x)^(-3/4)

Next, we substitute a = 0 into both the original function and its derivative:

f(0) = (1+2(0))^(1/4) = 1^(1/4) = 1

f'(0) = (1/2)(1+2(0))^(-3/4) = (1/2)(1)^(-3/4) = (1/2)

Now we can calculate the linear approximation L(x):

L(x) = f(0) + f'(0)(x-0)

= 1 + (1/2)x

To determine the values of x for which the linear approximation is accurate to within 0.1, we need to find the interval where |f(x) - L(x)| < 0.1.

|f(x) - L(x)| = |(1+2x)^(1/4) - (1 + (1/2)x)|

To simplify this expression, we can square both sides to eliminate the absolute value:

(f(x) - L(x))^2 < (0.1)^2

((1+2x)^(1/4) - (1 + (1/2)x))^2 < 0.01

By expanding and simplifying, we will get a quadratic inequality. However, it is difficult to solve analytically. Therefore, we will use numerical methods to approximate the solution.

We can plot the graph of the quadratic inequality using a graphing calculator or software and find the x-values where the graph is within 0.01 of the x-axis.

Alternatively, we can use an interval bisection method to narrow down the interval where the solutions lie. We can start with a large interval, such as [-10, 10], and continually divide it in half until we find a smaller interval containing the solutions accurate to within 0.1.

By using either of these methods, we can determine the values of x for which the linear approximation is accurate to within 0.1 and express the answers using interval notation rounded to three decimal places.