Question:

Find the intervals in which the function
\( f(x) = x^4 - 4x^3 + 4x^2 + 15 \) is increasing .

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Solution and Explanation

Step 1: For increasing function, we need \( f'(x) > 0 \).
 
Step 2: Differentiate the function:
\[ f'(x) = \frac{d}{dx}(x^4 - 4x^3 + 4x^2 + 15) = 4x^3 - 12x^2 + 8x \] Factor the derivative:
\[ f'(x) = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2) \] 
Step 3: Solve \( f'(x) > 0 \):
\[ 4x(x - 1)(x - 2) > 0 \] Critical points: \( x = 0, 1, 2 \) 
Step 4: Sign chart analysis:
 

  • Interval (-∞, 0): pick \( x = -1 \) 
    \( 4(-1)(-2)(-3) = -24 \) → Negative
  • Interval (0, 1): pick \( x = 0.5 \) 
    \( 4(0.5)(-0.5)(-1.5) = +1.5 \) → Positive
  • Interval (1, 2): pick \( x = 1.5 \) 
    \( 4(1.5)(0.5)(-0.5) = -1.5 \) → Negative
  • Interval (2, ∞): pick \( x = 3 \) 
    \( 4(3)(2)(1) = +24 \) → Positive

Conclusion:
\( f'(x) > 0 \) for intervals \( (0, 1) \) and \( (2, \infty) \)
Therefore, f(x) is strictly increasing in (0, 1) and (2, ∞)
Similarly, f(x) is decreasing in (−∞, 0) and (1, 2)

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