Question

Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), \[ \int_0^a f(x) \, dx = f(a), \] and given that \[ f(1) = 1, \quad f(16) = \frac{1}{8}, \] then \[ 16 - f^{-1}\left( \frac{1}{16} \right) \] is equal to:

Hint:

When dealing with functional equations involving integrals, use the properties of derivatives and the fundamental theorem of calculus to simplify and solve.

Answer 1

Step 1: The given information includes an integral condition \( \int_0^a f(x) \, dx = f(a) \), which implies a relationship between the function and its integral. 
Step 2: Differentiate both sides of the equation \( \int_0^a f(x) \, dx = f(a) \) with respect to \( a \). Using the fundamental theorem of calculus and the chain rule, we get: \[ f(a) = f'(a) \] This gives us an important condition for \( f \). 
Step 3: Now use the information about \( f(16) \) and \( f^{-1} \) to find the value of \( 16 - f^{-1}\left( \frac{1}{16} \right) \). Thus, the final value of \( 16 - f^{-1}\left( \frac{1}{16} \right) \) is found.