Question

The minimum value of the function \[ y = x^4 - 2x^2 + 1 \] in the interval \( \left[\frac{1}{2}, 2 \right] \) is:

• A. \( 0 \)
• B. \( 2 \)
• C. \( 8 \)
• D. \( 9 \)

Hint:

To find the minimum of a function in an interval, evaluate critical points and endpoints.

Answer 1

The Correct Option is A

Step 1: Finding the critical points.
Differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = 4x^3 - 4x \] Setting \( \frac{dy}{dx} = 0 \) for critical points: \[ 4x(x^2 - 1) = 0 \] \[ x(x - 1)(x + 1) = 0 \] \[ x = 0, 1, -1 \] Step 2: Evaluating within the given interval.
The interval is \( \left[\frac{1}{2}, 2\right] \), so we consider \( x = 1 \), \( x = \frac{1}{2} \), and \( x = 2 \). Computing function values: \[ y(1) = 1^4 - 2(1^2) + 1 = 0 \] \[ y\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 - 2\left(\frac{1}{2}\right)^2 + 1 = \frac{1}{16} - \frac{2}{4} + 1 = \frac{13}{16} \] \[ y(2) = 2^4 - 2(2^2) + 1 = 16 - 8 + 1 = 9 \] Step 3: Conclusion.
The minimum value in \( \left[\frac{1}{2}, 2\right] \) is \( 0 \) at \( x = 1 \). Thus, the correct answer is (A) \( 0 \).