Tangents and inverses Suppose L(x)=ax+b (with a≠0) is the equation of the line tangent to the graph of a one-to-one function f at (x0,y0). Also, suppose M(x)=cx+d is the equation of the line tangent to the graph of f^−1 at (y0,x0).


c. Prove that L^−1(x)=M(x).

Verified step by step guidance

1. Start by recalling the definition of the tangent line to a function at a point. For the function f at the point (x0, y0), the tangent line L(x) can be expressed as L(x) = f'(x0)(x - x0) + y0, where f'(x0) is the derivative of f at x0.
2. Next, consider the inverse function f^−1. The tangent line M(x) at the point (y0, x0) can be expressed similarly as M(x) = f^−1'(y0)(x - y0) + x0, where f^−1'(y0) is the derivative of the inverse function at y0.
3. Use the relationship between the derivatives of a function and its inverse. Specifically, recall that f'(x0) * f^−1'(y0) = 1, which implies that f^−1'(y0) = 1/f'(x0).
4. Substitute f^−1'(y0) into the equation for M(x) to express it in terms of f'(x0). This will help in establishing a connection between L(x) and M(x).
5. Finally, to prove that L^−1(x) = M(x), show that the inverse of the tangent line L(x) can be expressed in the same form as M(x) by manipulating the equations derived in the previous steps.