Suppose f is differentiable on (-∞,∞) and the equation of the line tangent to the graph of f at x = 2 is y = 5x -3. Use the linear approximation to f at x = 2 to approximate f(2.01).

검증된 단계별 안내

Identify that the tangent line equation y = 5x - 3 provides the slope of the tangent line, which is the derivative of f at x = 2. Therefore, f'(2) = 5.

Recognize that the linear approximation formula is f(x) ≈ f(a) + f'(a)(x - a), where a is the point of tangency, which in this case is x = 2.

Substitute the known values into the linear approximation formula: f(2.01) ≈ f(2) + f'(2)(2.01 - 2).

Use the tangent line equation to find f(2). Since the tangent line passes through the point (2, f(2)), substitute x = 2 into y = 5x - 3 to find f(2).

Calculate the approximation using the values obtained: f(2.01) ≈ f(2) + 5(0.01).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Differentiability

A function is differentiable at a point if it has a defined derivative at that point, meaning it has a tangent line. This implies that the function is continuous and smooth around that point. Differentiability is crucial for understanding how functions behave locally and is foundational for concepts like linear approximation.

Tangent Line

The tangent line to a function at a given point represents the best linear approximation of the function near that point. It is defined by the slope, which is the derivative at that point, and the point of tangency. In this case, the tangent line at x = 2 is given by the equation y = 5x - 3, indicating that the slope is 5.

Linear Approximation

Linear approximation uses the tangent line to estimate the value of a function near a specific point. The formula for linear approximation is f(x) ≈ f(a) + f'(a)(x - a), where a is the point of tangency. This method allows us to approximate f(2.01) using the known values of f(2) and its derivative at that point.