Question:
The function $f(x)=[x(x-2)]^{2}$ is increasing in the set}
The function $f(x)=[x(x-2)]^{2}$ is increasing in the set}
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Use the wavy curve method for inequalities involving polynomial derivatives.
Updated On: Jun 19, 2026
- $(-\infty,0)\cup(2,\infty)$
- $(-\infty,1)$
- $(1,2)$
- $(0,1) \cup(2,\infty)$
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The Correct Option is D
Solution and Explanation
Step 1: Formula
A function is increasing where $f'(x) > 0$.
Step 2: Analysis
- $f(x) = (x^2 - 2x)^2$. - $f'(x) = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$.
Step 3: Calculation
- Critical points: $x = 0, 1, 2$. - Interval test: - $(-\infty, 0) \implies (-)(-)(-) = (-)$ - $(0, 1) \implies (+)(-)(-) = (+)$ - $(1, 2) \implies (+)(+)(-) = (-)$ - $(2, \infty) \implies (+)(+)(+) = (+)$
Step 4: Conclusion
Function increases on $(0, 1) \cup (2, \infty)$. Final Answer: (D)
A function is increasing where $f'(x) > 0$.
Step 2: Analysis
- $f(x) = (x^2 - 2x)^2$. - $f'(x) = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$.
Step 3: Calculation
- Critical points: $x = 0, 1, 2$. - Interval test: - $(-\infty, 0) \implies (-)(-)(-) = (-)$ - $(0, 1) \implies (+)(-)(-) = (+)$ - $(1, 2) \implies (+)(+)(-) = (-)$ - $(2, \infty) \implies (+)(+)(+) = (+)$
Step 4: Conclusion
Function increases on $(0, 1) \cup (2, \infty)$. Final Answer: (D)
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