## Jan 11, 2018

See below

#### Explanation:

$f \left(x\right) = 3 + \frac{20}{{x}^{2} - 4}$

(a)

(i)

Clearly the point $A$ is on the y-axis, so $x = 0.$

For the y-coordinate find $f \left(0\right)$:

$f \left(0\right) = 3 + \frac{20}{{0}^{2} - 4} = 3 - 5 = - 2$

So $A$ is at $\left(0 , - 2\right)$.

(ii)

$f \left(x\right) = 3 + 20 {\left({x}^{2} - 4\right)}^{- 1}$

$\to f ' \left(x\right) = - 2 x \cdot 20 {\left({x}^{2} - 4\right)}^{-} 2$

$= 2 x \cdot \frac{- 20}{{x}^{2} - 4} ^ 2 = \frac{- 40 x}{{x}^{2} - 4} ^ 2$

$f ' \left(0\right) = \frac{- 40 \left(0\right)}{{\left(0\right)}^{2} - 4} ^ 2 = 0$

(b)

(i)

$f ' ' \left(x\right) = \frac{40 \left(3 {x}^{2} + 4\right)}{{x}^{2} - 4} ^ 3$

$f ' ' \left(0\right) = \frac{40 \left(3 {\left(0\right)}^{2} + 4\right)}{{\left(0\right)}^{2} - 4} ^ 3 = \frac{160}{-} 64 = - \frac{5}{2} < 0$

Since $f ' ' \left(0\right) < 0$ then the point must be a maximum.

(ii)

Set $f ' ' \left(x\right) = 0$ then:

$\frac{40 \left(3 {x}^{2} + 4\right)}{{x}^{2} - 4} ^ 3 = 0$

$\to 3 {x}^{2} + 4 = 0$

which clearly has no real roots. In order for a point of inflection to exist $f ' ' \left(x\right) = 0$ for some real value of $x$. But we has just shown that cannot happen.

(c)

From the graph we can clearly see horizontal asymptotes at $y = 3$ for very large $| x |$. If we look at the function:

$f \left(x\right) = 3 + \frac{20}{{x}^{2} - 4}$

The ${x}^{2}$ term on the denominator of the fraction makes the fractional term very small for very large |x| so we can say that:

$f \left(x\right) = 3 + \frac{20}{{x}^{2} - 4} \to 3$ as $x$ gets very large. Hence it behaves like a constant at $3$ as demonstrated on the plot.