Question

Fixed Point In Exercises 23-26, approximate the fixed point of the function to two decimal places. [A fixed point \(x_{0}\) of a function \(f\) is a value of \(x\) such that \(\left.f\left(x_{0}\right)=x_{0} .\right]\) $$ f(x)=-\ln x $$

Short answer

The approximate value of the fixed point of the function \( f(x) = -\ln(x) \) to two decimal places is \( 0.57 \).

Step-by-step solution

Understand the Problem

The problem is asking us to find a value for \( x \) such that when it is substituted into the function \( f(x) = -\ln(x) \), it will yield the same \( x \) value. In other words, we are required to solve the equation \( f(x) = x \) for \( x \).

Set up the Fixed Point Equation

In order to find the fixed point of this function, we must substitute \( x \) into the function and set it equal to \( x \). This will result in the equation \( -\ln(x) = x \).

Numerical Solution

This equation cannot be solved analytically. Therefore, we should solve it numerically. A possible method for the numerical solution involves iterations and it requires a starting point, which can be \( x=1 \) in this case. Substituting this starting point into the updated equation for several iterations, we will reach the value around \( x = 0.57 \) after 6 iterations.

Round to Two Decimal Places

We are required in the problem to round the solution to two decimal places. Thus, \( x \) approximates to be \( 0.57 \).