Question:
Let \(f(x) = x^3 - 3x^2 + 2\) be a function defined on \((-1, 3]\).
Which of the following statements is/are correct?
Let \(f(x) = x^3 - 3x^2 + 2\) be a function defined on \((-1, 3]\).
Which of the following statements is/are correct?
Which of the following statements is/are correct?
Show Hint
Find the critical points with the first derivative, classify them with the second derivative, and do not forget to check the domain endpoints.
Updated On: Jul 22, 2026
- \(f(x)\) has exactly two roots in \([-0.9, 0]\).
- \(f(x)\) has a minimum at 2 only.
- \(f(x)\) has maximum at 0 only.
- \(f(x)\) has a root at 1.
Show Solution
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The Correct Option is B, D
Solution and Explanation
Step 1: Find the critical points using the first derivative.
\[ f(x) = x^3 - 3x^2 + 2, \quad f'(x) = 3x^2 - 6x = 3x(x-2) \]
Setting \(f'(x)=0\) gives \(x=0\) and \(x=2\). Both lie inside the domain \((-1,3]\).
Step 2: Classify the critical points using the second derivative.
\[ f''(x) = 6x - 6 \]
At \(x=0\): \(f''(0) = -6 < 0\), so \(x=0\) is a local maximum, with \(f(0) = 2\).
At \(x=2\): \(f''(2) = 6 > 0\), so \(x=2\) is a local minimum, with \(f(2) = 8 - 12 + 2 = -2\).
Step 3: Find the roots of f(x).
Testing \(x=1\): \(f(1) = 1 - 3 + 2 = 0\), so \(x=1\) is a root, confirming option (D).
Dividing out \((x-1)\): \(x^3 - 3x^2 + 2 = (x-1)(x^2 - 2x - 2)\).
Solving \(x^2 - 2x - 2 = 0\) with the quadratic formula gives \(x = 1 \pm \sqrt{3}\), which is about \(2.732\) and \(-0.732\).
So the three roots of \(f(x)\) are approximately \(-0.732\), \(1\), and \(2.732\).
Step 4: Check option (A), roots in [-0.9, 0].
Of the three roots, only \(-0.732\) lies inside \([-0.9, 0]\). The roots at \(1\) and \(2.732\) fall outside this interval. So there is exactly one root there, not two, making (A) FALSE.
Step 5: Check option (B), minimum at 2 only.
On the domain \((-1,3]\), the point \(x=-1\) is excluded, so the value \(f(-1) = -1 - 3 + 2 = -2\) is only an approached limit, never actually attained. The local minimum value \(-2\) is genuinely reached only at \(x=2\). So (B) is TRUE.
Step 6: Check option (C), maximum at 0 only.
\(f(0) = 2\) is the local maximum. But check the right endpoint: \(f(3) = 27 - 27 + 2 = 2\), and \(x=3\) is included in the domain. So the value \(2\) is attained at both \(x=0\) and \(x=3\), meaning the maximum is not unique to \(x=0\). So (C) is FALSE.
Step 7: Check option (D), root at 1.
Already shown in Step 3, \(f(1) = 0\), so (D) is TRUE.
Final Answer:
Statements (B) and (D) are correct. \[ \boxed{\text{(B) and (D)}} \]
\[ f(x) = x^3 - 3x^2 + 2, \quad f'(x) = 3x^2 - 6x = 3x(x-2) \]
Setting \(f'(x)=0\) gives \(x=0\) and \(x=2\). Both lie inside the domain \((-1,3]\).
Step 2: Classify the critical points using the second derivative.
\[ f''(x) = 6x - 6 \]
At \(x=0\): \(f''(0) = -6 < 0\), so \(x=0\) is a local maximum, with \(f(0) = 2\).
At \(x=2\): \(f''(2) = 6 > 0\), so \(x=2\) is a local minimum, with \(f(2) = 8 - 12 + 2 = -2\).
Step 3: Find the roots of f(x).
Testing \(x=1\): \(f(1) = 1 - 3 + 2 = 0\), so \(x=1\) is a root, confirming option (D).
Dividing out \((x-1)\): \(x^3 - 3x^2 + 2 = (x-1)(x^2 - 2x - 2)\).
Solving \(x^2 - 2x - 2 = 0\) with the quadratic formula gives \(x = 1 \pm \sqrt{3}\), which is about \(2.732\) and \(-0.732\).
So the three roots of \(f(x)\) are approximately \(-0.732\), \(1\), and \(2.732\).
Step 4: Check option (A), roots in [-0.9, 0].
Of the three roots, only \(-0.732\) lies inside \([-0.9, 0]\). The roots at \(1\) and \(2.732\) fall outside this interval. So there is exactly one root there, not two, making (A) FALSE.
Step 5: Check option (B), minimum at 2 only.
On the domain \((-1,3]\), the point \(x=-1\) is excluded, so the value \(f(-1) = -1 - 3 + 2 = -2\) is only an approached limit, never actually attained. The local minimum value \(-2\) is genuinely reached only at \(x=2\). So (B) is TRUE.
Step 6: Check option (C), maximum at 0 only.
\(f(0) = 2\) is the local maximum. But check the right endpoint: \(f(3) = 27 - 27 + 2 = 2\), and \(x=3\) is included in the domain. So the value \(2\) is attained at both \(x=0\) and \(x=3\), meaning the maximum is not unique to \(x=0\). So (C) is FALSE.
Step 7: Check option (D), root at 1.
Already shown in Step 3, \(f(1) = 0\), so (D) is TRUE.
Final Answer:
Statements (B) and (D) are correct. \[ \boxed{\text{(B) and (D)}} \]
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