Melody

Thank you :)

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

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}

It looks like a standard question that you could find in any school.

There is nothing unusual about this question.

Graph them using desmos. And then answer for yourself

Very nicely answered Loki  

Here is my way, it is similar, just a bit different

\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0. \)

Think of the numerator and denominator separately. 

They both have to be positive OR they both have to be negative in order for the final answer to be positive.

If you let  y=x^2+x+3 

first notice it is a parabola.

Solve it for y=0 you will find there are no real answers, the discriminant under the square root sign is negative.

Since the leading coefficient is + it must be a concave up.

A concave up parabola with no real roots MUST always have a positive y value.

So I have discovered that the numerator (top) is always positive.

So the bottom must be positive too.

the denominator (bottom) is also a concave up parabola when set equal to y.

If there are any roots then the positive bit must be on either end of the roots (not in the middle)

Solving for 0=2x^2+x+6 gives  x=-2 and x=1.5

SO  the expression will be positive so for all values of x less than -2 or more than 1.5

Please post a link to your graph.  Graphs and algebra are learned together.

Good, I hope it is, you are too lazy to even lay your question out properly. 

And too lazy to search old answers.

I would have preferred that you were not answered at all but if the answer is wrong that is the next best thing.

(No disrespect intended to answering guest, I know you are just trying to help)

Why don't you share your solution with other site users?

The point ( x, 3.6) lies on a circle in the 2nd quadrant with diameter 8. What is the x-value?

Umm

Yes x can be any negative value.  See my graph.