Suppose f is differentiable on (-∞,∞) and f(5.01) - f(5) = 0.25.Use linear approximation to estimate the value of f'(5).

검증된 단계별 안내

First, recall the formula for linear approximation: f(x) ≈ f(a) + f'(a)(x - a). This formula is used to approximate the value of a function near a point a using the function's derivative at that point.

In this problem, we are given f(5.01) - f(5) = 0.25. This represents the change in the function value as x changes from 5 to 5.01.

Using the linear approximation formula, the change in the function value can be expressed as f(5.01) - f(5) ≈ f'(5)(5.01 - 5).

Substitute the given change in function value into the equation: 0.25 ≈ f'(5)(0.01).

Solve for f'(5) by dividing both sides of the equation by 0.01: f'(5) ≈ 0.25 / 0.01.

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주요 개념

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

Differentiability

A function is said to be differentiable at a point if it has a defined derivative at that point. This means that the function is smooth and continuous around that point, allowing for the calculation of the slope of the tangent line. Differentiability implies continuity, but not vice versa.

Linear Approximation

Linear approximation is a method used to estimate the value of a function near a given point using the tangent line at that point. The formula for linear approximation is f(x) ≈ f(a) + f'(a)(x - a), where 'a' is the point of tangency. This technique is particularly useful for estimating function values when the exact calculation is complex.

Derivative Interpretation

The derivative of a function at a point represents the instantaneous rate of change of the function with respect to its variable at that point. In practical terms, it can be interpreted as the slope of the tangent line to the function's graph at that point. Understanding this concept is crucial for applying linear approximation effectively.