Question:
The difference between the local extreme values of the function \(f(x) = 2x^3-15x^2+36x+40\) is ......
The difference between the local extreme values of the function \(f(x) = 2x^3-15x^2+36x+40\) is ......
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Find $f'=0$, then compute $f$ at both critical points.
Updated On: Oct 1, 2026
- \(4\)
- \(3\)
- \(2\)
- \(1\)
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The Correct Option is D
Solution and Explanation
Step 1: Critical points
\(f'(x)=6x^2-30x+36=6(x-2)(x-3)\), so \(x=2,3\).
Step 2: Classify
\(f''(x)=12x-30\). At \(x=2\), \(f''=-6<0\) (local max). At \(x=3\), \(f''=6>0\) (local min).
Step 3: Values
\(f(2)=16-60+72+40=68\) and \(f(3)=54-135+108+40=67\).
Difference \(=68-67=1\).
Final Answer:
The difference is \(1\), option (D). \[ \boxed{\text{(D) }1} \]
\(f'(x)=6x^2-30x+36=6(x-2)(x-3)\), so \(x=2,3\).
Step 2: Classify
\(f''(x)=12x-30\). At \(x=2\), \(f''=-6<0\) (local max). At \(x=3\), \(f''=6>0\) (local min).
Step 3: Values
\(f(2)=16-60+72+40=68\) and \(f(3)=54-135+108+40=67\).
Difference \(=68-67=1\).
Final Answer:
The difference is \(1\), option (D). \[ \boxed{\text{(D) }1} \]
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