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If $x$ is real, the maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ is?

Is it necessary that this function will attain maximum when the denominator will be minimum?

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The denominator is $$ \underbrace{x^2+2x+9 = (x^2+2x+1)+8}_\text{completing the square} = (x+1)^2 + 8 = (\text{a square}) + 8 $$ and that is always positive, never $0$. Therefore this function is everywhere continuous. As $x\to\pm\infty$, the function approaches $3$. Consequently, it must have a global maximum value unless it is always less than $3$.

If there is is a global maximum value, then, since the function is everywhere differentiable, the maximum must occur at a point where the derivative is $0$. If there are only finitely many points where the derivative is $0$, then find the value of the function at each of them. If the value is more than $3$ at one such point, then there is a maximum value, and you just have to pick the biggest value among those finitely many.

The derivative is a fraction, and a fraction is $0$ only if the numerator is $0$.

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Note that by division $\dfrac{3x^2+9x+17}{x^2+2x+9}=3+\dfrac{3x-10}{x^2+2x+9}$, so if the first expression has a maximum (or minimum) value it will occur when $\dfrac{3x-10}{x^2+2x+9}$ has its maximum (or minimum) value. The advantage of using $\dfrac{3x-10}{x^2+2x+9}$ is that the derivative is algebraically simpler.

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As the denominator is positive, we only need to find $M$ that allows equality in: $$3x^2+9x+17 \le M(x^2+2x+9)$$ $$\iff (M-3)x^2+(2M-9)x+(9M-17) \ge 0$$ For this to hold for large $|x|$, we need $M> 3$. Further, for equality, the LHS must touch the $X$ axis, so we need $$\triangle = 0 \implies (2M-9)^2=4(M-3)(9M-17) \implies M = \frac{35+\sqrt{241}}{16} \approx 3.158$$

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You need to solve:

$$\frac{\text{d}}{\text{d}x}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=0\Longleftrightarrow$$ $$\frac{47+x(20-3x)}{(9+x(2+x))^2}=0\Longleftrightarrow$$ $$47+x(20-3x)=0\Longleftrightarrow$$ $$x(20-3x)=-47\Longleftrightarrow$$ $$-3x^2+20x=-47\Longleftrightarrow$$ $$3x^2-20x=47$$

If you look to it graphicly you see that the maximum is about $8.5$ so finding the solutions of the equation we get:

$$x=\frac{10+\sqrt{241}}{3}\approx8.5081$$

So:

$$\text{max}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=\frac{35+\sqrt{241}}{16}\space\text{at}\space x=\frac{10+\sqrt{241}}{3}$$

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  • $\begingroup$ Why the downvote, can the one who did, explain it? Because I got the right answer!! $\endgroup$ – Jan Eerland Mar 3 '16 at 17:54
  • $\begingroup$ I think it's because it isn't complicated enough to use the d/dx thing :) $\endgroup$ – William Phoenix Sep 5 '16 at 9:18

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