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).

Verified step by step guidance

1. 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.
2. 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.
3. Substitute the known values into the linear approximation formula: f(2.01) ≈ f(2) + f'(2)(2.01 - 2).
4. 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).
5. Calculate the approximation using the values obtained: f(2.01) ≈ f(2) + 5(0.01).

Key Concepts

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.