InterviewSolution
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InterviewSolution
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Consider the function\({\rm{\;f\;}}\left( {\rm{x}} \right) = {{\rm{x}}^2} - {\rm{x}} - 2\). The maximum value of f (x) in the closed interval [- 4, 4] is1. 182. 103. 24. 4 |
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Answer» Correct Answer - Option 1 : 18 We have \({\rm{\;f\;}}\left( {\rm{x}} \right) = {{\rm{x}}^2} - {\rm{x}} - 2\) \(\begin{array}{l} {{\rm{f}}^{'}}\left( {\rm{x}} \right) = 2{\rm{x}} - 1 = 0 \to {\rm{x}} = \frac{1}{2}\\ {{\rm{f}}^{''}}\left( {\rm{x}} \right) = 2 \end{array}\) Since\({\rm{f'}}\left( {\rm{x}} \right) = 2 > 0\), thus \({\rm{x}} = \frac{1}{2}\) is minimum point. The maximum value in closed interval [-4, 4] will be at x = -4 or x = 4 Now maximum value \( = {\rm{max}}\left[ {{\rm{f\;}}\left( { - {\rm{\;}}4} \right),{\rm{\;f\;}}\left( 4 \right)} \right] = {\rm{max}}\left( {18,10} \right) = 18\) |
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