# Thread: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

1. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

Hmm...alternatively you can let $x = \frac{\cot \theta}{\sqrt{3}}$, you should still obtain the same thing.

2. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

It's okay, since mostly, we use diffentiation while tri sub. is not a always useful method instead. Don't feel bad for not thinking for tri method, it's okay

3. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

Daigo has just taken Algebra. Soon, i think the suggests should somewhat easier to understand; not calculus or trigonometry, but basic Algebra.

You should estimate the points on the graph that are max. and min.

4. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

Aha! I thought there would be an non-calculus, non-trig, algebraic solution to this problem.

Let k be the maximal value of $\frac{2x}{3x^2 + 1}$ so that $\frac{2x}{3x^2 + 1} = k$. Cross-multiplying, we obtain

$3kx^2 - 2x + k = 0 \Rightarrow x = \frac{2 \pm \sqrt{4 - 4(3k)(k)}}{6k} = \frac{2 \pm \sqrt{4 - 12k^2}}{6k}$. We want k to be as large as possible subject to the constraint that x is real. This implies that

$4 - 12k^2 \ge 0 \Rightarrow k^2 \le \frac{1}{3}$. The largest possible value of k is $\sqrt{\frac{1}{3}}$.

Furthermore, since $y = \frac{2x}{3x^2 + 1}$ is an odd function, the smallest possible value of k is $-\sqrt{\frac{1}{3}}$.

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