Show that $ h(x) = \frac{2x}{x^2 + 5} - \frac{4}{x^2 + 5} - \frac{18}{(x^2 + 5)(x + 2)} $, for $ x \geq 0 $ - Edexcel - A-Level Maths Pure - Question 28 - 2013 - Paper 1

We can use the quotient rule to find the derivative of $h(x)$:

If $u = 2(x^2 - 13)$ and $v = (x^2 + 5)(x + 2)$, then:

$h'(x) = \frac{u'v - uv'}{v^2}$

Calculating $u'$ and $v'$:

• $u' = 4x$.
• We apply the product rule to find $v'$:

$v = (x^2 + 5)(x + 2)$ $v' = (2x)(x + 2) + (x^2 + 5)(1)$ $= 2x^2 + 4x + x^2 + 5$ $= 3x^2 + 4x + 5$

Thus, applying these in the quotient rule gives us: $h'(x) = \frac{4x((x^2 + 5)(x + 2)) - 2(x^2 - 13)(3x^2 + 4x + 5)}{((x^2 + 5)(x + 2))^2}$

This expression can be simplified to find the simplest form.

To find the range of $h(x)$, we need to examine its behavior as $x \to 0$ and as $x \to \infty$.

1. As $x \to 0$: $h(0) = \frac{2(0) - 26}{(0 + 5)(2)} = \frac{-26}{10} = -2.6.$

2. As $x \to \infty$: The leading terms of the numerator and denominator dominate: $h(x) \sim \frac{2x^2}{x^3} = 0.$ Hence, $h(x) \to 0$ as $x \to \infty$.

3. Finding Critical Points: By setting $h'(x) = 0$, we can find the local maximum (details derived from previous steps show that) occurs at: $x = \sqrt{5}.$ Plugging $\sqrt{5}$ into $h(x)$ gives: $h(\sqrt{5}) = \frac{2(5) - 26}{(5 + 5)(\sqrt{5} + 2)} = \frac{-16}{10 + 2\sqrt{5}}.$

So the range of $h(x)$ is: $\left(-\infty, \frac{h(\sqrt{5}) \text{max}}{10 + 2\sqrt{5}} \right)$. Thus: $$ Range(h) = \left(-\infty, 0 \right) $.