Understanding the Given Function

The function provided is $f(x) = (|x| - |x – 1|)^2$. This function involves absolute values, $|x|$ and $|x-1|$. To analyze the function properly, we need to consider different cases based on the values of $x$ where the expressions inside the absolute values change sign. The critical points are $x=0$ (for $|x|$) and $x=1$ (for $|x-1|$).

Let's define the function $f(x)$ piecewise:

• Case 1: $x < 0$
  In this case, $x$ is negative, so $|x| = -x$. Also, $x-1$ is negative (since $x < 0$, $x-1$ is even smaller), so $|x-1| = -(x-1) = 1-x$.
  $|x| - |x-1| = (-x) - (1-x) = -x - 1 + x = -1$.
  So, $f(x) = (-1)^2 = 1$ for $x < 0$.
• Case 2: $0 \le x < 1$
  In this case, $x$ is non-negative, so $|x| = x$. Also, $x-1$ is negative, so $|x-1| = -(x-1) = 1-x$.
  $|x| - |x-1| = x - (1-x) = x - 1 + x = 2x - 1$.
  So, $f(x) = (2x - 1)^2 = 4x^2 - 4x + 1$ for $0 \le x < 1$.
• Case 3: $x \ge 1$
  In this case, $x$ is greater than or equal to 1, so $|x| = x$. Also, $x-1$ is non-negative, so $|x-1| = x-1$.
  $|x| - |x-1| = x - (x-1) = x - x + 1 = 1$.
  So, $f(x) = (1)^2 = 1$ for $x \ge 1$.

Combining these cases, the function $f(x)$ is defined as:

$$f(x) = \begin{cases} 1 & \text{if } x < 0 \\ (2x-1)^2 & \text{if } 0 \le x < 1 \\ 1 & \text{if } x \ge 1 \end{cases}$$

Evaluating Statement 1: f(-2) = f(5)

Let's evaluate the function at $x = -2$ and $x = 5$ using the piecewise definition.

• For $x = -2$: Since $-2 < 0$, we use the first case, $f(x) = 1$. So, $f(-2) = 1$.
• For $x = 5$: Since $5 \ge 1$, we use the third case, $f(x) = 1$. So, $f(5) = 1$.

Comparing the values, we have $f(-2) = 1$ and $f(5) = 1$. Therefore, $f(-2) = f(5)$.

Statement 1 is correct.

Evaluating Statement 2: f”(-2) + f”(0.5) + f”(3) = 4

To evaluate this statement, we first need to find the second derivative of the function $f(x)$ in each interval where it is smooth. The function might not be differentiable or twice differentiable at the points where the definition changes ($x=0$ and $x=1$), but the points we need to evaluate $f''$ at (-2, 0.5, 3) are away from these critical points.

Let's find the first derivative, $f'(x)$, and the second derivative, $f''(x)$, for each interval:

• For $x < 0$:
  $f(x) = 1$
  $f'(x) = \frac{d}{dx}(1) = 0$
  $f''(x) = \frac{d}{dx}(0) = 0$
• For $0 < x < 1$:
  $f(x) = (2x-1)^2 = 4x^2 - 4x + 1$
  $f'(x) = \frac{d}{dx}(4x^2 - 4x + 1) = 8x - 4$
  $f''(x) = \frac{d}{dx}(8x - 4) = 8$
• For $x > 1$:
  $f(x) = 1$
  $f'(x) = \frac{d}{dx}(1) = 0$
  $f''(x) = \frac{d}{dx}(0) = 0$

So, the second derivative $f''(x)$ is:

$$f''(x) = \begin{cases} 0 & \text{if } x < 0 \\ 8 & \text{if } 0 < x < 1 \\ 0 & \text{if } x > 1 \end{cases}$$

Now, let's evaluate $f''$ at the points specified in the statement:

• For $x = -2$: Since $-2 < 0$, $f''(-2) = 0$.
• For $x = 0.5$: Since $0 < 0.5 < 1$, $f''(0.5) = 8$.
• For $x = 3$: Since $3 > 1$, $f''(3) = 0$.

Now, let's calculate the sum $f''(-2) + f''(0.5) + f''(3)$:

Sum $= 0 + 8 + 0 = 8$.

The statement claims that the sum is 4. However, we calculated the sum to be 8. Therefore, $f''(-2) + f''(0.5) + f''(3) = 4$ is false.

Statement 2 is incorrect.

Conclusion on Statements

Based on our analysis:

• Statement 1: $f(-2) = f(5)$ is Correct.
• Statement 2: $f''(-2) + f''(0.5) + f''(3) = 4$ is Incorrect.

Therefore, only statement 1 is correct.

Revision Table: Key Steps

Additional Information on Absolute Value Functions and Derivatives

Functions involving absolute values, like $|x|$ or $|x-c|$, are often defined piecewise. The points where the expression inside the absolute value is zero (like $x=0$ for $|x|$ or $x=c$ for $|x-c|$) are called critical points. At these critical points, the graph of the absolute value function has a sharp corner, and the derivative does not exist there.

When dealing with a function like $f(x) = (|x| - |x-1|)^2$, the critical points from the absolute values are $x=0$ and $x=1$. The function needs to be analyzed in the intervals defined by these points: $(-\infty, 0)$, $[0, 1)$, and $[1, \infty)$.

Derivatives of piecewise functions are found by differentiating the expression for each interval separately. However, one must be careful about the derivative at the boundary points. A function is differentiable at a boundary point only if the left-hand derivative and the right-hand derivative exist and are equal at that point. For the second derivative, the same logic applies to the first derivative function.

In this specific problem, the points where we needed to evaluate the second derivative (-2, 0.5, 3) were in the "smooth" intervals where the function's form was simple (either a constant or a quadratic). Therefore, we could find the second derivative in those intervals directly and evaluate it at the given points.

The function $f(x) = (|x| - |x-1|)^2$ is interesting because $|x| - |x-1|$ simplifies to a constant (-1 or 1) outside the interval $[0, 1)$, leading to $f(x)=1$ in those regions. Inside the interval $[0, 1)$, it is a quadratic $(2x-1)^2$. This structure explains why the second derivative is 0 outside $(0, 1)$ and a constant (8) inside $(0, 1)$.