# Can someone explain this word problem?

Oct 22, 2016

$x \in \left(- 3 , - \frac{1}{5}\right) \cup \left(4 , \infty\right)$

#### Explanation:

Let's try to convert this into a more mathematical representation.

The positions of the planes are determined in relation to the top of a 10-metre pole: planes to the left and below the top of the pole are given negative x- and y- values and planes to the right and above the top of the pole are given positive x- and y- values.

The "x- and y- values" talk here is a big clue. This is describing a cartesian plane, with the origin (the $\left(0 , 0\right)$ point) at the top of the pole. A position on a cartesian plane is a point, meaning we are representing the planes as points.

The path of Tyler's plane is modelled by f(x) = ..."
The path of Kym's plane is modelled by g(x) = ..."

As a point travels along a path on the cartesian plane, it forms a curve. As each plane flies along a path, and we are representing these planes as points, we can represent their flight paths as curves. In this case, we are given $y = f \left(x\right)$ and $y = g \left(x\right)$ as polynomials equations which represent their paths.

Determine when f(x) > g(x).

After all that, it turns out we didn't really need to worry about all the setup to solve the problem. All we needed was $f$ and $g$. If we care about the physical interpretation, we are finding out values for $x$ result in the $y -$ coordinate of Tyler's plane being greater than the $y -$ coordinate of Kym's plane, that is, at what distances Tyler's plane flew higher.

But all that aside, we can solve the inequality with only the last three sentences, which boil down to:

$7 {x}^{3} - 11 {x}^{2} - 22 x + 8 > - 3 {x}^{3} - 3 {x}^{2} + 100 x + 32$
Solve for $x$.

While we're here, let's do that.

$7 {x}^{3} - 11 {x}^{2} - 22 x + 8 > - 3 {x}^{3} - 3 {x}^{2} + 100 x + 32$

$\implies 10 {x}^{3} - 8 {x}^{2} - 122 x - 24 > 0$

$\implies 5 {x}^{3} - 4 {x}^{2} - 61 x - 12 > 0$

Letting $h \left(x\right) = 5 {x}^{3} - 4 {x}^{2} - 61 x - 12$, we can test for where $h \left(x\right) = 0$ using the rational roots theorem.

Our potential candidates are of the form $\pm \frac{p}{q}$ where $p$ is a factor of $12$ and $q$ is a factor of $5$. Doing so, we find that $h \left(4\right) = h \left(- 3\right) = h \left(- \frac{1}{5}\right) = 0$ (note that after finding one of these, say $4$, we could have divided by $x - 4$ and then found the roots of the remaining quadratic using the quadratic formula).

So we have $\left(x - 4\right) , \left(x + 3\right) ,$ and $\left(x + \frac{1}{5}\right)$ as factors of $h \left(x\right)$. We also need the leading coefficient to be $5$. Multiplying by $5$, we get

$h \left(x\right) = 5 \left(x - 4\right) \left(x + 3\right) \left(x + \frac{1}{5}\right) = \left(x - 4\right) \left(x + 3\right) \left(5 x + 1\right)$

$\implies \left(x - 4\right) \left(x + 3\right) \left(5 x + 1\right) > 0$

This inequality holds true when either all of the factors are positive, or two are negative and one is positive (the product of two negatives is a positive). If we check what happens to the factors as we traverse the real line, we get:

$\left(- \infty , - 3\right) : - - -$
$\left(- 3 , - \frac{1}{5}\right) : - + -$
$\left(- \frac{1}{5} , 4\right) : - + +$
$\left(4 , \infty\right) : + + +$

So, the solution to the inequality, and thus to the problem, is the union of the second and fourth intervals:

$x \in \left(- 3 , - \frac{1}{5}\right) \cup \left(4 , \infty\right)$