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While solving a problem, I came across this function: $$y=\frac{12+x+0.02x^2}{3+0.1x}$$ It is a linear function except it is discontinuous at a single point. When I plotted it in wolfram alpha, it suggested the alternate form of $y=0.2x+4$ with the condition that $x\neq-30$. The problem would have been much easier to solve using the alternate form given that $x=-30$ is not a physical solution.

My question would be, how did wolfram determine this? Obviously when I look at the graph I can come up with the same linear function, but is there a way to mathematically "simply away the -30 case" or something to get the alternate form just from the original equation? (without plotting)

Ryan
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  • How did you decide that the function was linear? Simply looking at the graph can be misleading. For example, the function for $\bar z$ here looks very much like a straight line when plotted, and even when you zoom in the curve is quite subtle, but if you compute the differences in its values for equally-spaced values of $h$, it becomes obvious that it’s not really linear. – amd Apr 17 '18 at 22:16

2 Answers2

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$$y=\frac{12+x+0.02x^2}{3+0.1x}=\frac{600+50x+x^2}{150+5x}=\frac{(20+x)(30+x)}{5(30+x)}=\frac{20+x}{5}=4+0.2x$$

where dividing through by $30+x$ requires $x \neq -30$

glowstonetrees
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This is just polynomial division.

You have $A=0.02x^2+x+12$ and $B=0.1x+3$

and try to find $Q,R$ such that $A=BQ+R$ with $\deg(R)<\deg(B)$

  • So you start with higher power coefficients : $0.02x^2$ and $0.1x$ this gives $Q=0.2x+\cdots$

$\require{cancel}A-0.2x\times B = (0.02x^2+x+12)-0.2x(0.1x+3)=\cancel{0.02x^2}+x+12-\cancel{0.02x^2}-0.6x=0.4x+12$

  • And you continue comparing higher power coefficients : $0.4x$ and $0.1x$ this gives $Q=0.2x+4$

$A-0.2xB-4B=0.4x+12-4(0.1x+3)=\cancel{0.4x}+\cancel{12}-\cancel{0.4x}-\cancel{12}=0$

Thus $R=0$ here, the division is exact.

zwim
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