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-1
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Find the range of$$f(x) = \frac{3x + 4}{x^2 + 3x + 5}$$
as $$x$$ varies over all real numbers.

Apr 26, 2020

#1
+21957
+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.

Apr 26, 2020
#2
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hi geno3141 thanks for responding. I tried your method but it's telling me that your answer is wrong

Guest Apr 26, 2020
#3
+111547
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Did you try graphing it using Desmos?

Melody  Apr 26, 2020
#4
+1

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

Guest Apr 26, 2020
#5
+111547
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Melody  Apr 26, 2020
#6
+1

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

Guest Apr 27, 2020
#8
+111547
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Thank you :)

Melody  Apr 27, 2020
#7
+111547
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$$f(x) = \frac{3x + 4}{x^2 + 3x + 5}$$

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  $$f'(x)=0$$

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

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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}\\