Find the set of values taken by 1\(9x^2+12x+7) for real values of X .

* In the first part of the question they asked me to make out the complete square of the quadratic formula alone with being under 1 . It was
(3x+2)^2+3 .

Question

Find the set of values taken by 1\(9x^2+12x+7) for real values of X . 

* In the first part of the question they asked me to make out the complete square of the quadratic formula alone with being under 1 . It was 
(3x+2)^2+3 .

ikram002p · Answer

9x^2+12x+7 $\neq$ 0

find value that makes it 0 , then domain would be R-|zeros of it |

anonymous · Answer

hey is the answer is to be given in the perspective of calculus

anonymous · Answer

I guess that basically the book is asking the range of

$$f(x)= \frac{ 1 }{ 9x^2 + 12x + 7 }$$
WHat's yar book answer?

anonymous · Answer

@Lama97

ikram002p · Answer

oh i thought it wont the domain :P

anonymous · Answer

its 

0<f(x) , or equal 1\3

anonymous · Answer

ikram I just thought the same way

anonymous · Answer

oh i meant less or equal to 1\3

ikram002p · Answer

:) then u need to find max which would be 1/3
and note that the function >0 , since 9x^2+12x+7 is positive

ikram002p · Answer

do you know how to find max ?

anonymous · Answer

max point ?  vertex right ?

ikram002p · Answer

first find derivative of the function , f'(x)
maximum point or minimum point would be when
f'(x) =0

anonymous · Answer

I think that it wants the possible solution for all the x on the no. line

SO therefore we have to prove that f(x) > 0

So we have to fing the maximma(or maximum value of f(x0

So I guess to solve this we have to differentiate the EQN.

SO let :
$$h(x) = 9x^2+12x+7$$

y = h(x)

SO :
We know that:

$$\frac{ d }{ dx }x^n = nx^{n-1}$$
$$\frac{ d }{ dx }h(x) = 18x+12 $$

Then h'(x) = 18x + 12(which is at min or max when h'(x) = 0)

SO:

If 18x+12 = 0 then x =-2/3(This  is minimma I guess)

Therefore I think that this point becomes a maximum in f(x)('cuz f(x) = 1/h(x)).

Therefore  the maximum of f(x) is f(-2/3) = 1/3 (just plug the number in the f(x)..Lol)

Isn't  g(x) is at a minimum for x=-2/3 and has the value of 3 and all other values of g(x) are greater than this. 

SO this means that f(x) is always +ve but >1/3 (>0 and=0 at infinity)

SO the answer is:

$$f(x)> 0$$

Or

$$f(x) = 1/3$$

ikram002p · Answer

f(-2/3)=1/3
:D

anonymous · Answer

@Lama97

anonymous · Answer

guys you both did  a great job , to whom should I give the medal :P

ikram002p · Answer

to @Adjax  :)
he explained it :P
i only gives hint xD

anonymous · Answer

well yes he explained it in details. but you both really deserve it. thanks ikram :) .. 
Adjax my great appreciation :)