# How to do these question?

May 24, 2018

#### Explanation:

Part (a)

$g \left(x\right) = {\left(x - a\right)}^{2} \left(x - b\right)$

$g ' \left(x\right) = 2 \left(x - a\right) \left(x - b\right) + {\left(x - a\right)}^{2} \left(1\right)$

$= \left(x - a\right) \left(3 x - \left(a + 2 b\right)\right)$

$g ' \left(x\right) = 0$ at $x = a$ and at $x = \frac{a + 2 b}{3}$

The stationary points are

$\left(a , 0\right)$ and $\left(\frac{a + 2 b}{3} , f \left(\frac{a + 2 b}{3}\right)\right) = \left(\frac{a + 2 b}{3} , \frac{4 {\left(a - b\right)}^{3}}{27}\right)$

Part (b) cannot be read and part (d) cannot be answered without knowing what $p$ is.

Part (d)

IF we interpret "there are two solutions" to mean "there are exactly two distinct solutions", and we assume that $a \ne b$, then we have
$p = 0$ which occurs at $\left(a , 0\right)$ and $\left(b , 0\right)$

or $p = \frac{4 {\left(a - b\right)}^{3}}{27}$ which occurs at $x = \frac{a + 2 b}{3}$ and at another value of $x$
That other value of $x$ may be found by using the fact that we know one solution (in fact a solution of multiplicity 2) of the equation $g \left(x\right) = \frac{4 {\left(a - b\right)}^{3}}{27}$. If nothing better occurs to us, we can do the division to factor $g \left(x\right) - \frac{4 {\left(a - b\right)}^{3}}{27}$