Question

The function f(x) = 2x - \(|x - x^2|\) is

• A. continuous at x = 1
• B. discontinuous at x = 1
• C. not defined at x = 1
• D. discontinuous at x = 0

Hint:

Absolute value functions $|g(x)|$ are continuous wherever $g(x)$ is continuous.

Answer 1

The Correct Option is A

To determine whether the function \( f(x) = 2x - |x - x^2| \) is continuous at \( x = 1 \), we need to examine the behavior of the function around this point. A function is continuous at a point if the following three conditions are satisfied:

1. The function is defined at the point; that is, \( f(1) \) exists.
2. The limit of the function as \( x \) approaches the point exists; that is, \( \lim_{x \to 1} f(x) \) exists.
3. The value of the limit equals the value of the function at that point; that is, \( \lim_{x \to 1} f(x) = f(1) \).

Let's verify each of these conditions:

Step 1: Calculate \( f(1) \)

Substitute \( x = 1 \) into the function:

\(f(1) = 2(1) - |1 - 1^2| = 2 - |1 - 1| = 2 - 0 = 2\)

Step 2: Calculate \( \lim_{x \to 1} f(x) \)

Since \( f(x) = 2x - |x - x^2| \), let's compute the limit from both sides:

For \( x < 1 \) (left-hand limit):
\( |x - x^2| = -(x - x^2) = x^2 - x \) (since \( x - x^2 \) is negative for \( x < 1 \))

• \(f(x) = 2x - (x^2 - x) = 3x - x^2\)
• For \( x > 1 \) (right-hand limit):
   

\(|x - x^2| = x - x^2\) (since \( x - x^2 \) is positive or zero for \( x > 1 \))

• \(f(x) = 2x - (x - x^2) = 2x - x + x^2 = x + x^2\)

Thus, both limits as \( x \) approaches 1 are:

Left-hand limit:

• \(\lim_{x \to 1^-} f(x) = 3(1) - 1^2 = 3 - 1 = 2\)

Right-hand limit:

• \(\lim_{x \to 1^+} f(x) = 1 + 1^2 = 1 + 1 = 2\)

Step 3: Verify Continuity

Since both the left-hand limit and right-hand limit equal 2, we have:

\(\lim_{x \to 1} f(x) = 2\)

Now, comparing with \( f(1) = 2 \), we conclude:

The limit \( \lim_{x \to 1} f(x) = f(1) \), thus the function is continuous at \( x = 1 \).

Therefore, the correct answer is: continuous at \( x = 1 \).