Complete the square and find the maximum value of f(x)=1/(x^2+2x+2).

Question

anonymous · Answer

The answer is 1 but idk how to solve it

anonymous · Answer

i assume it means complete the square in the denominator
it is not necessary to do that to answer this question, but we can do it if you like
there is a much easier way

anonymous · Answer

$$f(x)=\frac{1}{x^2+2x+2}=\frac{1}{x^2+2x+1+1}=\frac{1}{(x+1)^2+1}$$

anonymous · Answer

now the the square has been completed, we can reason that the smallest the denominator can be is $1$ which it is if $x=-1$
then since the smallest the denominator can be is 1, the largest the entire function can be is $$f(-1)=\frac{1}{1}=1$$

anonymous · Answer

we could have done the same work by minimizing 
$$x^2+2x+2$$ which has a minimum at the vertex
the first coordinate of the vertex is $-\frac{b}{2a}=-\frac{2}{2}=-1$ and the second coordinate of the vertex is $(-1)^2+2(-1)+2=1$

anonymous · Answer

thank you!

anonymous · Answer

yw