# 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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### 1 Answer

`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.*