Question:

For what value of \(x\), the function \(2x^3+3x^2-36x+10\) has minimum?

Show Hint

For maximum or minimum, first solve \(f'(x)=0\). Then use \(f''(x)>0\) for minimum and \(f''(x)<0\) for maximum.
  • \(-2\)
  • \(-3\)
  • \(2\)
  • \(1\)
Show Solution
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The Correct Option is C

Solution and Explanation

Let the function be \[ f(x)=2x^3+3x^2-36x+10. \] To find the point where the function has a minimum, first find the critical points by differentiating. \[ f'(x)=\frac{d}{dx}(2x^3+3x^2-36x+10). \] \[ f'(x)=6x^2+6x-36. \] Take \(6\) common: \[ f'(x)=6(x^2+x-6). \] Factorize: \[ x^2+x-6=(x+3)(x-2). \] So, \[ f'(x)=6(x+3)(x-2). \] For critical points: \[ f'(x)=0. \] \[ 6(x+3)(x-2)=0. \] Therefore, \[ x=-3 \] or \[ x=2. \] Now use the second derivative test. \[ f''(x)=\frac{d}{dx}(6x^2+6x-36). \] \[ f''(x)=12x+6. \] At \(x=2\): \[ f''(2)=12(2)+6=24+6=30. \] Since \[ f''(2)>0, \] the function has a minimum at \[ x=2. \] Hence, the required value of \(x\) is \[ 2. \]
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