Let `f(x) = x^2+2x+9; x in RR`
find the greatest value of `1/f(x)` , giving the value of x for which it is attained.
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`f(x) = x^2+2x+9`
By completing square;
`f(x) = (x+1)^2+8`
`1/f(x) = 1/((x+1)^2+8)`
We know that `(x+1)^2>=0` always.
Hence `(x+1)^2+8>0 ` always.
When `f(x)` is increasing then `1/f(x)` decreases. On the other hand maximum `1/f(x)` is obtained when `f(x)` is minimum.
Since `(x+1)^2>=0` always minimum f(x) is obtained when `(x+1)^2=0` or when x = -1.
minimum `f(x) = 0+8 = 8`
Maximum `1/f(x) = 8`
So the maximum of `1/f(x)` is 8 and it is obtained when x = -1.
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Please accept the correction.
Minimum `f(x) = 8`
Hence Maximum `1/f(x) = 1/8` and it is obtained when x = -1.
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