This is a continuation of the previous question.

What is the Linear Approximation of
𝑓
(
𝑥
)
=
tan
⁡
𝑥
at
𝑥
=
𝜋
4
?

Question

This is a continuation of the previous question.

What is the Linear Approximation of 
𝑓
(
𝑥
)
=
tan
⁡
𝑥
 at 
𝑥
=
𝜋
4
 ?

BlackTom AI · Accepted Answer

We start by restating the problem to focus on what needs to be found. The task is to obtain the linear approximation (the tangent-line approximation) of f(x) = tan x at x = π/4, and then compare with the given options.

First, recall the formula for the linear (Taylor) approximation at a = π/4: L(x) = f(a) + f'(a)(x − a).
- Compute f(a): f(π/4) = tan(π/4) = 1.
- Compute f'(x): the derivative of tan x is sec^2 x. Therefore f'(x) = sec^2 x.
- Evaluate f'(a): f'(π/4) = sec^2(π/4). Since cos(π/4) = √2/2, sec(π/4) = 1/cos(π/4) = 2/√2 = √2, and sec^2(π/4) = (√2)^2 = 2.
Putting it together: L(x) = 1 + 2(x − π/4) = 1 + 2x − π/2 = 2x − π/2 + 1.

Now examine each option:
- Option A: L(x) = x/2 + π/4. This form expands to 0.5x + 0.785…, which does not match the derived linear expression 2x − π/2 + 1. The coefficient of x is wrong (0.5 vs 2), and the constant term is also not aligned with 1 − π/2.
- Option B: L(x) = 2x − π/2. This matches the linear part in x but is missing the constant term +1. Without +1, the value at x = π/4 would be 2(π/4) − π/2 = π/2 − π/2 = 0, whereas the true tangent line at that point must pass through (π/4, 1). The discrepancy shows this is insufficient as the full linear approximation.
- Option C: L(x) = 2x − π/4 + 1. Here the x-coefficient 2 is correct, and the constant term includes +1, but the constant offset −π/4 differs from the needed −π/2. This shifts the line vertically relative to the correct tangent line, so it does not match the true L(x).
- Option D: L(x) = x − π/4 + 1. This has the wrong slope (coefficient of x is 1, not 2) and an offset that does not align with the tangent at π/4. Therefore it cannot be the linear approximation.
- Option E: L(x) = 2x − π/2 + 1. This expression exactly matches the derived linear approximation L(x) = 1 + 2(x − π/4) = 2x − π/2 + 1. The slope is 2, the intercept is consistent with f(π/4) = 1, and the line passes through the point (π/4, 1).

In summary, each option’s validity hinges on both correct slope and correct intercept to satisfy L(π/4) = f(π/4) and L'(x) = f'(x) at the point. Options with inconsistent slopes fail to match the derivative at the expansion point, while those with incorrect intercepts fail to pass through the function value at the expansion point. The option that aligns with both the computed function value at the expansion point and the derivative value yields the exact linear approximation.

This is a continuation of the previous question. What is the Linear Approximation of 𝑓 ( 𝑥 ) = tan ⁡ 𝑥 at 𝑥 = 𝜋 4 ?Single choice

Similar Questions

at 的线性近似值为：提示：at 的线性近似值为：。

The linear approximation of 𝑓 ( 𝑥 ) = 5 𝑥 − 4 2 𝑥 2 + 1 at 𝑥 = − 3 is: Hint: The linear approximation of 𝑦 = 𝑓 ( 𝑥 ) at 𝑥 = 𝑥 0 is: 𝐿 ( 𝑥 ) = 𝑓 ′ ( 𝑥 0 ) ( 𝑥 − 𝑥 0 ) + 𝑓 ( 𝑥 0 ) .

Suppose you want to use a Linear Approximation to approximate 1 25.2 without a calculator. Which of the following functions and 𝑥 -values would be best to use?

Consider the Linear Approximation of the function 𝑓 ( 𝑥 ) at 𝑥 = 𝑎 . Which of the following is the best description of what the Linear Approximation does?

Suppose f ( x , t ) = e − 2 t sin ⁡ ( x + 3 t ) . Which of the following is a good approximation of the value of f ( 2.02 , − 0.03 ) ? (Hint: first find the differential d f at the point ( 2 , 0 ) .)

Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is473754\frac{73}{75}.52125\frac{2}{12}.42325.4\frac{23}{25}.497100.4\frac{97}{100}.41315.4\frac{13}{15}.

Question at position 3 By using differentials, an approximation of 1233\sqrt[3]{123} is497100.4\frac{97}{100}.41315.4\frac{13}{15}.42325.4\frac{23}{25}.52125\frac{2}{12}.473754\frac{73}{75}.

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler