A student is searching for a solution to the equation $f(x) = 0$ - AQA - A-Level Maths Pure - Question 2 - 2020 - Paper 1

To determine which function does not adhere to the conclusion that there is a root between $-1$ and $1$ based on the evaluations, we analyze each option:

1. For $f(x) = \frac{1}{x}$:

   • $f(-1) = -1$ (negative)
   • $f(1) = 1$ (positive)
   • Correct conclusion, as there is a root at $x = 0$, which is between -1 and 1.

2. For $f(x) = x$:

   • $f(-1) = -1$ (negative)
   • $f(1) = 1$ (positive)
   • Correct conclusion, as there is a root at $x = 0$, which is between -1 and 1.

3. For $f(x) = x^3$:

   • $f(-1) = -1$ (negative)
   • $f(1) = 1$ (positive)
   • Correct conclusion, as there is a root at $x = 0$, which is between -1 and 1.

4. For $f(x) = \frac{2x + 1}{x + 2}$:

   • $f(-1) = 0$ (zero)
   • $f(1) = 1$ (positive)
   • The conclusion is especially maintained since $f(-1) = 0$, indicating that there is a root at $x = -1$, but does not show any change of sign needed for $x$ between $(0, 1)$.

Thus, the function for which the conclusion is incorrect is $f(x) = \frac{1}{x}$, as $f(x)$ is undefined at $x = 0$, leading to no change of sign. Therefore, the final answer is:

Circle your answer: $f(x) = \frac{1}{x}$