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Find the range of\(f(x) = \frac{3x + 4}{x^2 + 3x + 5}\)
as \(x\) varies over all real numbers.

Give your answer in interval notation.

 Apr 26, 2020
 #1
avatar+20788 
+1

If you find f'(x)  =  ( -3x2 - 8x + 3 ) / ( x2 + 3x + 5 )2

set it equal to zero and solve, you find the maximum to be 1/3 and the minimum to be -3.

Answer:  [ -3, 1/3 ]

 Apr 26, 2020
 #2
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0

hi geno3141 thanks for responding. I tried your method but it's telling me that your answer is wrong

Guest Apr 26, 2020
 #3
avatar+109486 
+1

Did you try graphing it using Desmos?

Melody  Apr 26, 2020
 #4
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+1

Yes but I would like to know how to solve this with algebra

Guest Apr 26, 2020
 #5
avatar+109486 
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Please post a link to your graph.  Graphs and algebra are learned together.

Melody  Apr 26, 2020
 #6
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+1

https://www.desmos.com/calculator/zurnxjuzkv 

Guest Apr 27, 2020
 #8
avatar+109486 
0

Thank you :)

Melody  Apr 27, 2020
 #7
avatar+109486 
+1

\(f(x) = \frac{3x + 4}{x^2 + 3x + 5} \)

 

If I look at top and bottom separately I can see that the bottom would be a concave up parabola with no roots.

Therefore,for all real values of x the denominator will always be positive.

 

Find the turning points. This is where  \(f'(x)=0\)

 

\(f(x) = \frac{3(x^2 + 3x + 5)-(2x+3)(3x + 4)}{(x^2 + 3x + 5)^2}\\ f(x) = \frac{-3x^2-8x+3}{(x^2 + 3x + 5)^2}\\ f'(x)=0\quad when\quad -3x^2-8x+3=0\\ \text{That happens when } x=-3,\;and \;\;when\;x=\frac{1}{3}\)

 

 

-----------

I only just realized that Geno had done that much and you showed no sign of understanding that his question was incomplete and a little wrong.

 

Do you understand the graph Desmos drew?    Can you tell me the answer from that graph?

You will need to use a little algebra as well but you do not have to do ALL the algebra to get the correct answer from the graph.

 

HINTS:

You need to find the exact max and min.

If you do the whole thing algebraically you would also need to get the y value as x approaches + and - infinity.  You can use L'hopital's rule to do that.  BUT you can see the answer from the graph anyway.

 

Come back with what you discover.

 

 

 

 

 

LaTex:

f(x) = \frac{3(x^2 + 3x + 5)-(2x+3)(3x + 4)}{(x^2 + 3x + 5)^2}\\
f(x) = \frac{-3x^2-8x+3}{(x^2 + 3x + 5)^2}\\
f'(x)=0\quad when\quad -3x^2-8x+3=0\\
\text{That happens when } x=-3,\;and \;\;when\;x=\frac{1}{3}

 Apr 27, 2020
edited by Melody  Apr 27, 2020

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