The minimum value of the function f(x) = x^4 - 8x^2 + 16 is:

Practice Questions

Q1
The minimum value of the function f(x) = x^4 - 8x^2 + 16 is:
  1. 0
  2. 1
  3. 2
  4. 4

Questions & Step-by-Step Solutions

The minimum value of the function f(x) = x^4 - 8x^2 + 16 is:
  • Step 1: Write down the function we want to analyze: f(x) = x^4 - 8x^2 + 16.
  • Step 2: Find the derivative of the function, which tells us the slope of the function. The derivative is f'(x) = 4x^3 - 16x.
  • Step 3: Set the derivative equal to zero to find the critical points: 4x^3 - 16x = 0.
  • Step 4: Factor the equation: 4x(x^2 - 4) = 0.
  • Step 5: Solve for x by setting each factor to zero: 4x = 0 gives x = 0, and x^2 - 4 = 0 gives x = ±2.
  • Step 6: Now we have three critical points: x = 0, x = 2, and x = -2.
  • Step 7: Evaluate the original function f(x) at each critical point: f(0) = 16, f(2) = 0, and f(-2) = 0.
  • Step 8: Compare the values obtained: 16, 0, and 0. The minimum value is the smallest of these, which is 0.
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