Question

If \( f(x) = \begin{cases} x^2 & \text{for } x < 0 \\ 5x - 3 & \text{for } 0 \leq x \leq 2 \\ x^2 + 1 & \text{for } x > 2 \end{cases} \), then the positive value of \( x \) for which \( f(x) = 2 \) is:

• A. \( \frac{3}{5} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{4} \)
• D. 1
• E. 0

Hint:

When solving for values in a piecewise function, ensure to consider the domain restrictions of each piece to identify viable solutions correctly.

Answer 1

The Correct Option is D

We need to find the value of \( x \) within the specified ranges where \( f(x) = 2 \). 
- For \( x < 0 \), \( f(x) = x^2 \): Setting \( x^2 = 2 \) gives \( x = \pm \sqrt{2} \). However, only negative \( x \) values are applicable in this range, which does not contribute to the positive solution. 
- For \( 0 \leq x \leq 2 \), \( f(x) = 5x - 3 \): Solving \( 5x - 3 = 2 \) yields: \[ 5x = 5 \implies x = 1 \] \( x = 1 \) is within the interval \( [0, 2] \) and satisfies \( f(x) = 2 \). 
- For \( x > 2 \), \( f(x) = x^2 + 1 \): Solving \( x^2 + 1 = 2 \) results in: \[ x^2 = 1 \implies x = \pm 1 \] Only \( x = 1 \) fits this equation, but \( x \) must be greater than 2 in this range, so \( x = 1 \) is not valid here. 
Thus, the correct and only positive \( x \) value for which \( f(x) = 2 \) under the constraints provided by the piecewise function is \( x = 1 \). 
Therefore, the correct answer is Option D.