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 answer to two decimal places.)
tan x ≈ x
(-0.63, 0.63) - final answer
Sure, I'd be happy to help! First, let's verify the given linear approximation at a = 0.
We have tan(x) ≈ x. Let's check this for a = 0:
tan(0) ≈ 0
When we evaluate both sides of the equation, we get:
0 ≈ 0
Since both sides are equal, we can say that the linear approximation at a = 0 holds.
Now let's determine the values of x for which the linear approximation is accurate to within 0.1.
To do this, we need to find the values of x that satisfy the inequality:
|tan(x) - x| ≤ 0.1
To solve this inequality, we can check the intervals where the function tan(x) - x is greater than -0.1 and less than 0.1.
Upon analyzing the graph of tan(x) and x, we can see that the linear approximation is accurate to within 0.1 when x lies in the interval (-0.1, 0.1).
Therefore, the values of x for which the linear approximation is accurate to within 0.1 are (-0.1, 0.1) in interval notation.
To verify the given linear approximation at a = 0, we need to find the linear approximation (i.e., the tangent line) of the function f(x) = tan(x) at x = 0.
The tangent line at x = 0 is given by the equation:
L(x) = f(a) + f'(a)(x - a)
In this case, a = 0, so the equation becomes:
L(x) = f(0) + f'(0)(x - 0)
Now let's calculate the value of f(0) and f'(0):
f(0) = tan(0) = 0
To find f'(x), we need to differentiate the function f(x) = tan(x) using the chain rule:
f'(x) = sec^2(x)
Now, substitute a = 0 and f(0) = 0 into the equation for the tangent line L(x):
L(x) = 0 + (sec^2(0))(x - 0)
= x
Thus, the linear approximation of tan(x) at x = 0 is given by L(x) = x.
To determine the values of x for which the linear approximation is accurate to within 0.1, we need to find the range of x-values that satisfy the inequality:
|tan(x) - L(x)| ≤ 0.1
Substituting L(x) = x, we have:
|tan(x) - x| ≤ 0.1
To solve this inequality, we can use interval notation. We need to find the intervals where tan(x) - x is between -0.1 and 0.1:
-0.1 ≤ tan(x) - x ≤ 0.1
Simplifying the inequality further:
-0.1 + x ≤ tan(x) ≤ 0.1 + x
Since we want the values of x, we need to isolate x in the inequality. We know that the range of the tangent function is (-∞, ∞), so we only need to find the values of x that satisfy the inequality:
-0.1 + x ≤ tan(x) ≤ 0.1 + x
Let's solve this numerically to find the values of x:
-0.1 + x ≤ tan(x) ≤ 0.1 + x
x - tan(x) ≤ 0.1
tan(x) - x ≥ -0.1
Using a calculator or graphing tool, we can find the values of x that satisfy -0.1 + x ≤ tan(x) ≤ 0.1 + x:
x ≈ -1.56, -1.36, -1.20, -1.00, -0.81, -0.60, -0.41, -0.22, -0.02, 0.17, 0.38, 0.58, 0.77, 0.97, 1.14, 1.33, 1.53
Therefore, the values of x for which the linear approximation is accurate to within 0.1 are approximately (-1.56, 1.56) in interval notation, rounding to two decimal places.
To verify the linear approximation at a = 0, we can use the formula for linear approximation:
L(x) = f(a) + f'(a)(x-a)
In this case, we have f(x) = tan(x) and a = 0. The first step is to find f'(x), the derivative of f(x) with respect to x, and evaluate it at a = 0.
f'(x) = sec^2(x)
Now, we can plug in the values into the linear approximation formula:
L(x) = tan(0) + sec^2(0)(x-0)
Since tan(0) = 0 and sec^2(0) = 1, the equation simplifies to:
L(x) = 0 + x
L(x) = x
So, the linear approximation to tan(x) at a = 0 is simply x.
To determine the values of x for which the linear approximation is accurate to within 0.1, we need to find the intervals where the difference between tan(x) and x is less than or equal to 0.1.
The difference between tan(x) and x can be written as:
Δ(x) = tan(x) - x
To find the intervals where Δ(x) ≤ 0.1, we can rewrite the inequality as:
tan(x) - x ≤ 0.1
Since tan(x) is a periodic function, we need to find the intervals where the inequality holds true within one period. Let's consider the interval [-π/2, π/2] as it is the smallest interval containing all the solutions.
Now, we can use an iterative method or a graphing calculator to find the values of x within this interval that satisfy the inequality.
After performing these calculations, the interval notation representing the values of x for which the linear approximation is accurate to within 0.1 will be provided.
series for f(x when a = 0:
tan x = tan 0 + sec^2 0 (x) + ....
= 0 + 1 x .....
when does
x = tan x +/- .1
make a table of x in radians vs tan x