| For any real number x, the maximum value of \[4-6x-{{x}^{2}}\] |
A) 4
B) 7
C) 9
D) 13
Correct Answer: D
Solution :
| Let the given equation be represented as |
| \[f(x)=4-6x-{{x}^{2}}\] |
| Now, differentiating above function w.r.t x, we get |
| \[f'(x)=-\,6-2x\] |
| For value of x put \[f'(x)=0\] |
| \[-\,6-2x=0\] |
| \[\therefore \] \[x=-\,3\] |
| For maximum value, we take \[f'(x)\] |
| \[f'(x)=-\,2\] |
| Since, value of \[f'(x)\] is negative. |
| So, \[f(x)\] is maximum at \[x=-\,3.\] |
| Putting in\[x=-\,3\] in \[f(x),\]we get |
| \[f(-\,3)=4-(6)\,\,(-\,3)-{{(-\,3)}^{2}}\] |
| \[=4+18-9=13\] |
| So, the maximum value of \[=4-6x-{{x}^{2}}\]is 13. |
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