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.

Guest Apr 26, 2020

#1**+1 **

If you find f'(x) = ( -3x^{2} - 8x + 3 ) / ( x^{2} + 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 ]

geno3141 Apr 26, 2020

#7**+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 \(f'(x)=0\)

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

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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}\\

f'(x)=0\quad when\quad -3x^2-8x+3=0\\

\text{That happens when } x=-3,\;and \;\;when\;x=\frac{1}{3}

Melody Apr 27, 2020