Better Students Ask More Questions.
Let `f(x) = x^2+2x+9; x in RR` find the greatest value of `1/f(x)` , giving the value...
1 Answer | add yours
`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.
Posted by jeew-m on September 17, 2013 at 7:54 PM (Answer #1)
1 Reply | Hide Replies ▲
Join to answer this question
Join a community of thousands of dedicated teachers and students.