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.
+23246 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 ]
+118608 \(f(x) = \frac{3x + 4}{x^2 + 3x + 5} \)
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}
