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Find the range of the function (x^2 +14x+9)/(x^2+2x+3)
as x varies over all real numbers.

 Aug 17, 2020
 #1
avatar+110835 
+3

This was my approach

 

\(y=\frac{(x^2 +14x+9)}{(x^2+2x+3)}\\ y=\frac{(x^2 +2x+3+12x+6)}{(x^2+2x+3)}\\ y=1+6*\frac{(2x+1)}{(x^2+2x+3)}\\ \text{I note that the denominator is positive for all real values of x}\\ \text{As x tends to }\pm\infty \text{ y tends to +1}\\ y'=6*\frac{(x^2+2x+3)(2)-(2x+2)(2x+1)}{(x^2+2x+3)^2}\\ y'=0 \quad when \quad \\ (x^2+2x+3)(2)-(2x+2)(2x+1)=0\\ 2x^2+4x+6-(4x^2+6x+2)=0\\ -2x^2-2x+4=0\\ x^2+x-2=0\\ (x+2)(x-1)=0\\ x=1\;\;or \;\;x=-2\\ \)

When x=1 y=4

When x=-2, y=-5

 

I now have enough info to know that the graph looks something like this:

 

 

 

 

\(\bf\text{So the range of this function is   }   -5 \le x\le 4\)

 

 

 

 

 

LaTex:

y=\frac{(x^2 +14x+9)}{(x^2+2x+3)}\\
y=\frac{(x^2 +2x+3+12x+6)}{(x^2+2x+3)}\\
y=1+6*\frac{(2x+1)}{(x^2+2x+3)}\\
\text{I note that the denominator is positive for all real values of x}\\
\text{As x tends to }\pm\infty \text{  y tends to +1}\\
y'=6*\frac{(x^2+2x+3)(2)-(2x+2)(2x+1)}{(x^2+2x+3)^2}\\
y'=0 \quad when \quad \\
(x^2+2x+3)(2)-(2x+2)(2x+1)=0\\
2x^2+4x+6-(4x^2+6x+2)=0\\
-2x^2-2x+4=0\\
x^2+x-2=0\\
(x+2)(x-1)=0\\
x=1\;\;or \;\;x=-2\\

 Aug 18, 2020
edited by Melody  Aug 18, 2020
 #2
avatar+1035 
+4

I did it differently, but Melody's answer is still very good. (Or maybe they are the same, and I just did not see...)

The first thing to do, as Melody did, is to set f(x) to y. This gives us:

y=(x^2 +14x+9)/(x^2+2x+3).

Now, we can multiply both sides by x^2+2x+3, and get:

y(x^2+2x+3)=x^2 +14x+9.

Now, we must solve for the discriminant:

x^2 + 14x + 9=yx^2+2yx+3y

x^2 + 14x + 9-(yx^2+2yx+3y)=0

(1-y)x^2 + (14-2y)x + 9-3y=0      <---- regrouping to get the coefficients

(14-y)^2-4(1-y)(9-3y)=196-56y+4y-(36-48y+12y^2)=-8y^2-8y+160      <--------discriminant formula

Now that we got our discriminant, we can say:

-8y^2-8y+160>=0   <------- or else it would be imaginary, which is no good

y^2+y-20 <= 0

(y+5)(y-4) <= 0

We see that for any number between -5 and 4, it works, so our range is [-5, 4]

 

:)

 Aug 18, 2020
 #3
avatar+110835 
+1

Hi  ilorty,

 

For this specific question your way is MUCH better than mine.

I have only been introduced to your method very recently, perhaps by you, and it will take me a while to integrate it into my own memory and use.

 

Thanks very much for reminding me  laugh 

Melody  Aug 18, 2020
 #4
avatar+1035 
+4

Your answer was very great too! Although I am a bit confused on what the purple line means...

ilorty  Aug 18, 2020
edited by ilorty  Aug 18, 2020
 #5
avatar+110835 
+1

The purple line is the asymptote.

I treated it as a graphing exercise.

Melody  Aug 18, 2020
 #6
avatar+1035 
+4

Oh~that makes a lot more sense. Thanks!

ilorty  Aug 18, 2020

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