Question

The greatest value of
\[\frac{4}{4x^2+4x+9}\]
is-

Ans.1/2
method please!!!!

Answer 1

firstly complete the square for 4x^2+4x+9

Answer 2

??

Answer 3

\[\large 4x^2+4x+9=(2x+1)^2+8\]
am i right?

Answer 4

yes

Answer 5

well now use this
\[(2x+1)^2+8\ge 8 \\ since\\ (2x+1)^2\ge0\]
for any real number x

Answer 6

what about the numerator?

Answer 7

\[\huge a>b \\ then \\ \huge \frac{1}{a}<\frac{1}{b}\]

Answer 8

what will u get?

Answer 9

i don't know , you solve ahead.

Answer 10

\[(2x+1)^2+8\ge8 \ \ then \ \ \frac{1}{(2x+1)^2+8} \le \frac{1}{8}\]

Answer 11

multiply it by 4 and u will get the answer

Answer 12

A fraction of constant numerator is maximum when the denominator is minimum.
As @mukushla is trying to explain to you that the denominator is minimum when
x=-1/2.
You can verify that the fraction is equal to 1/2 when x=-1/2

Answer 13

Also you can do this problem using derivatives
\[
f(x)=\frac{4}{4
x^2+4 x+9}\\
f'(x)=-\frac{16 (2 x+1)}{\left(4
x^2+4 x+9\right)^2}
\]
You can deduce that f has a maximum at x =-1/2