Question

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/((1 + 9x)^4) ≈ 1 − 36x

Answer 1

Answer:

Part 1)

See Below.

Part 2)

$\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)$

Step-by-step explanation:

Part 1)

The linear approximation L for a function f at the point x = a is given by:

$\displaystyle L \approx f'(a)(x-a) + f(a)$

We want to verify that the expression:

$1-36x$

Is the linear approximation for the function:

$\displaystyle f(x) = \frac{1}{(1+9x)^4}$

At x = 0.

So, find f'(x). We can use the chain rule:

$\displaystyle f'(x) = -4(1+9x)^{-4-1}\cdot (9)$

Simplify. Hence:

$\displaystyle f'(x) = -\frac{36}{(1+9x)^{5}}$

Then the slope of the linear approximation at x = 0 will be:

$\displaystyle f'(1) = -\frac{36}{(1+9(0))^5} = -36$

And the value of the function at x = 0 is:

$\displaystyle f(0) = \frac{1}{(1+9(0))^4} = 1$

Thus, the linear approximation will be:

$\displaystyle L = (-36)(x-(0)) + 1 = 1 - 36x$

Hence verified.

Part B)

We want to determine the values of x for which the linear approximation L is accurate to within 0.1.

In other words:

$\displaystyle \left| f(x) - L(x) \right | \leq 0.1$

By definition:

$\displaystyle -0.1\leq f(x) - L(x) \leq 0.1$

Therefore:

$\displaystyle -0.1 \leq \left(\frac{1}{(1+9x)^4} \right) - (1-36x) \leq 0.1$

We can solve this by using a graphing calculator. Please refer to the graph shown below.

We can see that the inequality is true (i.e. the graph is between y = 0.1 and y = -0.1) for x values between -0.179 and -0.178 as well as -0.010 and 0.012.

In interval notation:

$\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)$