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.

MAT135H5_F25_ALL SECTIONS 4.2 Preparation Check

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

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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}.

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