In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:
d. Find the equation for the tangent line to g at the point (f(x_0), x_0) located symmetrically across the 45° line y=x (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line.
67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3

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Identify the given function as \( y = f(x) = \sqrt{3x - 2} \) and the point \( x_0 = 3 \). The function \( g \) is the inverse of \( f \), so \( g = f^{-1} \). The point on \( g \) corresponding to \( x_0 \) is \( (f(x_0), x_0) \).

Calculate \( f(x_0) \) by substituting \( x_0 = 3 \) into \( f(x) \): \( f(3) = \sqrt{3 \cdot 3 - 2} \). This gives the coordinates of the point on \( g \) as \( (f(3), 3) \).

Use Theorem 1, which states that if \( g = f^{-1} \), then the derivative of \( g \) at \( f(x_0) \) is \( g'(f(x_0)) = \frac{1}{f'(x_0)} \). So, first find \( f'(x) \) by differentiating \( f(x) = \sqrt{3x - 2} \).

Evaluate \( f'(x) \) at \( x_0 = 3 \) to find \( f'(3) \). Then compute \( g'(f(3)) = \frac{1}{f'(3)} \), which is the slope of the tangent line to \( g \) at the point \( (f(3), 3) \).

Write the equation of the tangent line to \( g \) at \( (f(3), 3) \) using the point-slope form: \[ y - y_1 = m (x - x_1) \] where \( (x_1, y_1) = (f(3), 3) \) and \( m = g'(f(3)) \). This line is symmetric to the tangent line of \( f \) at \( x_0 = 3 \) across the line \( y = x \).

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주요 개념

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

Inverse Functions and Their Graphs

An inverse function reverses the roles of inputs and outputs of the original function, swapping x and y. Graphically, the inverse function's graph is the reflection of the original function's graph across the line y = x (the 45° line). Understanding this symmetry is crucial for locating points like (f(x₀), x₀) on the inverse function.

Derivative of an Inverse Function (Theorem 1)

Theorem 1 states that if f is differentiable and invertible at x₀ with f'(x₀) ≠ 0, then the derivative of its inverse g at y₀ = f(x₀) is g'(y₀) = 1 / f'(x₀). This relationship allows us to find the slope of the tangent line to the inverse function using the derivative of the original function.

Equation of a Tangent Line

The tangent line to a function at a point provides the best linear approximation near that point. Its equation is y - y₁ = m(x - x₁), where m is the slope (derivative at the point) and (x₁, y₁) is the point of tangency. For the inverse function, the point and slope must be carefully identified using the inverse relationship.