For the function \( f(x) = x^3 \), \( x = 0 \) is a point of:

Question

Let \( f \) be a polynomial function such that \[ f(x^2+1)=x^4+5x^2+2,\quad \text{for all } x\in\mathbb{R}. \] Then \[ \int_0^3 f(x)\,dx \] is equal to:

Answer 1

Step 1: Function Definition
The function provided is f(x) = x³, a continuous and differentiable cubic function.

Step 2: First Derivative Calculation
The first derivative is f'(x) = 3x². Setting f'(x) = 0 yields x = 0 as a critical point.

Step 3: Second Derivative Calculation
The second derivative is f''(x) = 6x. At the critical point x = 0, f''(0) = 0, indicating the second derivative test is inconclusive.

Step 4: Concavity Analysis
The sign of f''(x) around x = 0 is analyzed: f''(x)<0 for x<0 (concave down) and f''(x)>0 for x>0 (concave up). This shows a change in concavity at x = 0.

Step 5: Conclusion
A change in concavity at x = 0 signifies an inflection point, where the curve's curvature changes but it is not a local extremum.

Final Answer: (D) inflexion

For the function \( f(x) = x^3 \), \( x = 0 \) is a point of:

The Correct Option is D

Solution and Explanation

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