How to graphically solve $x^3-x+1=0$ if the graph of $y=f(x)=x^2+x+(3/x)$ is given?

Asked Aug 07 '19 at 08:32

Active Aug 07 '19 at 18:45

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Problem:

The graph of $y=f(x)=x^2+x+(3/x)$ is given on a piece of paper. We are asked to find the solution of $x^3-x+1=0$ only by using the given graph.

Attempt:

Find another "simpler" graph $g(x)=ax^2+bx+c$ such that $f(x)=g(x)$ leads to the same equation $x^3-x+1=0$. By equating the coefficients, I get $a=0$, $b=1$ and $c=1$. Unfortunately, the constants are not equal.

$f(x)=g(x)$ produces $x^3-x+3=0$ while the equation in question is $x^3-x+1=0$. Assume the question is valid (or no typos), then how to solve it?

Is there any simpler $g(x)$ than $g(x)=ax^2+bx+c$?

edited Aug 07 '19 at 11:32

asked Aug 07 '19 at 08:32

Display Name

2,715

I don't understand if the division applies only to the last summand or to entire expression. Is it $f(x) = x^2+x+\frac3x$? – dfnu Aug 07 '19 at 11:30
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@dfnu: Last term only. – Display Name Aug 07 '19 at 11:30
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Well, honestly, it does not seem like a great approach, but you could draw on the same graph the function $g(x) = 1+x+\frac2x$ and see where it intersects $f(x)$... – dfnu Aug 07 '19 at 11:39
Where does this problem come from? Have you devised it? – Piquito Aug 07 '19 at 12:46
@Piquito: An unofficial IGCSE mocking test from someone who did not want to be famous. :-) – Display Name Aug 07 '19 at 12:50
@MoneyOrientedProgrammer: "Mocking test" seems to me the real serious thing in this problem. I'm wrong? – Piquito Aug 07 '19 at 13:05
@Piquito: Or a "mock exam". I don't understand. :-) – Display Name Aug 07 '19 at 13:35
@MoneyOrientedProgrammer: My English is weak. I am sorry. Regards. – Piquito Aug 07 '19 at 14:12

How to graphically solve $x^3-x+1=0$ if the graph of $y=f(x)=x^2+x+(3/x)$ is given?

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