Which values can $x(x-3)/(x-4)$ take?

Which values can $x(x-3)/(x-4)$ take?

Question

If $y=\frac{x(x-3)}{x-4}$ and $x$ is real show that $y$ cannot take any value between $+1$ and $+9$.

Sketch the curve $y=\dfrac{x(x-3)}{x-4}$.

Using the same axes, sketch the curve $y=5/x$ and find its intersection with the line $x=4$.

Hence prove that the equation $\frac{5}{x}=\frac{x(x-3)}{x-4}$ has no positive root.

Suggestion

Could we write $y=\frac{x(x-3)}{x-4}$ as a quadratic in $x$?

Can we now use the discriminant of this?

Or alternatively, maybe calculus might help us here?

Can we find the stationary points?

Here are some thoughts to consider while sketching any graph:

• Where does the graph intersect the axes?

• Where is the graph positive (above the $x$-axis) and where is it negative?

• Does the graph have any asymptotes, horizontal, vertical or diagonal?

• How does the graph behave as $x\to\pm\infty$?

• Does the graph have any stationary points? If so, where, and what is their nature (maximum, minimum, point of inflection)?

Using the same axes, sketch the curve $y=\dfrac{5}{x}$ and find its intersection with the line $x=4$.

Hence prove that the equation $\frac{5}{x}=\frac{x(x-3)}{x-4}$ has no positive root.

What is going on in the two separate regions $x<4$ and $x>4$ here?

Solution

Approach 1: Using the Discriminant

Multiplying through by the denominator and rearranging, we get $x^2-(y+3)x+4y=0.$

Since

$x$

is real, this must have real roots. Therefore its discriminant is non-negative, that is

$$(y+3)^2-16y\ge0.$$

Solving this inequality, we find

$$\begin{align*} &&(y+3)^2-16y &{}\geq 0&&\quad \\ \iff \quad&&y^2+6y+9-16y &{}\geq 0 \\ \iff \quad&&y^2-10y+9 &{}\geq 0 \\ \iff \quad&&(y-9)(y-1) &{}\geq 0, \end{align*}$$

so we must have $y \geq 9$ or $y\leq1$, as we can see from this sketch.

Approach 2: Using calculus

We can find the stationary points of the function by differentiating.

Using the Quotient Rule, we have

$$\begin{align*} \frac{dy}{dx}&{}=\frac{(2x-3)(x-4)-x(x-3)}{(x-4)^2}\\ &{}=\frac{x^2-8x+12}{(x-4)^2}\\ &{}=\frac{(x-2)(x-6)}{(x-4)^2}. \end{align*}$$

So there are stationary points at $(2,1)$ and $(6,9)$.

Around $x=2$, we have the points $(1,\frac{2}{3})$ and $(3,0)$, so $(2,1)$ is a local maximum.

Around $x=6$, we have the points $(5,10)$ and $(7,\frac{28}{3})$, so $(6,9)$ is a local minimum.

Combining these with the observation that $x=4$ is an asymptote, we can see that $y\le1$ when $x<4$ and $y\ge9$ when $x>4$.

We have an asymptote at $x=4$, since the denominator is zero there.

To identify where the function is positive and negative, we note that it is the product and quotient of the three terms $x$, $x-3$and $x-4$.

To find the intercepts, we note that when $x=0$, $y=0$. And when $y=0$, we require $x(x-3)=0$, so $x=0$ or $x=3$.

To find out how the function behaves as $x\rightarrow \pm \infty$, we perform the division: $\frac{x(x-3)}{x-4}=\frac{x^2-3x}{x-4}=x+1+\frac{4}{x-4},$ so as $x\to\pm\infty$, $y\approx x+1$, and we have an oblique asymptote.

We found above the stationary points at $(2,1)$ and $(6,9)$.

We can now put the above together to sketch the graph.

It is interesting to note that our graph appears to have rotational symmetry about the intersection of the two asymptotes, at $(4,5)$.

note that $(2,1)$ rotates to $(6,9)$, for example. How could we prove that the graph really does have rotational symmetry?

Using the same axes, sketch the curve $y=\dfrac{5}{x}$ and find its intersection with the line $x=4$.

The intersection of the curves $y=\dfrac{5}{x}$ and $x=4$ is $\left(4,\dfrac{5}{4}\right)$.

From the graph, it appears that solving $y = \dfrac{5}{x}$ with $y=\dfrac{x(x-3)}{x-4}$ simultaneously gives one negative root, but no positive ones.

For $x<4$, we have $\dfrac{5}{x}>\dfrac{5}{4}$, while we know from above that $\dfrac{x(x-3)}{x-4}\le 1<\dfrac{5}{4}$, so there is no root with $0<x<4$.

Similarly, for $x>4$, we have $\dfrac{5}{x}<\dfrac{5}{4}$, while $\dfrac{x(x-3)}{x-4}\ge 9>\dfrac{5}{4}$, so there is no root with $x>4$.

Clearly $x = 4$ is not a root.

Hence the equation $\dfrac{5}{x}=\dfrac{x(x-3)}{x-4}$ has no positive roots.