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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