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How can I draw the graph of the function: $f(x)=x^3-3x+1$,that has 3 solutions, to determine the number of solutions contained in the interval: $[-2,2]$?

Hagen von Eitzen
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wonderingdev
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    What sort of math do you have available to use? Obviously finding the solutions to the cubic is probably not allowed, right? ;) – apnorton Jan 29 '14 at 17:13
  • Some brief graphing tips: (1) You can determine the number and approximate location of the $x$-intercepts by looking at where $y = x^3$ and $y = 3x - 1$ intersect. (Note that each of these can be easily sketched by hand.) (2) The dominate term is $x^{3},$ which tells you what the graph looks like for "large" negative values of $x$ and "large" positive values of $x.$ (3) When $x$ is close to $0,$ the graph resembles $y = -3x + 1.$ (4) Every cubic has two turning points or no turning points, and by now you should know which is the case for this cubic. (continued) – Dave L. Renfro Jan 29 '14 at 17:19
  • (continuation) (5) Note that $x^3 - 3x + 1 = x(x^2 - 3) + 1,$ so you can try graphing $y = x(x^2 - 3)$ (easy using sign charts; note there are three $x$-intercepts: $x = 0, \pm \sqrt{3}$), and then shift the graph vertically upward by $1$ unit. – Dave L. Renfro Jan 29 '14 at 17:22

2 Answers2

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A good idea to get some overview is to compute a few values, i.e. $f(-2)= -8+6+1=-1$, $f(-1)= -1+3+1=3$, $f(0) = 1$, $f(1) = 1-3+1=-1$, $f(2) = 8-6+1=2$. The graph has to pas through the points $(-1,-1)$, $(-1,3)$, $(0,0)$, $(1,-1)$ and $(2,2)$. By the IVT this already shows that there is (at least) one root in $\left]-2,-1\right[$, one in $\left]0,1\right[$ and one in $\left]1,2\right[$. Since there cannot be more than three roots anyway, we have shown that all three roots are in $[-2,2]$.

Hagen von Eitzen
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This is a simple polynomial, so it shouldn't be very difficult.

  • First, find out its roots, that is the value where $f(x) = 0$. This can be done by trial-and-error, or algebraically solving it. Once you find out the roots, you know where the graph will cut the x-axis!

  • Then differentiate the function to get another polynomial representing the function's slope at every point. Then find out the roots to $f'(x) = 0$, which will show you where the function has maximas and minimas.
    Now substituting the x-values over here, you can find the value of these extremas.

  • Plugging in intermediate values between roots will also tell you where the graph has what sign.

  • Finally, put in $x \rightarrow±\infty$ and see where the graph will end up.

You can verify your graph on sites like Wolfram Alpha or GraphSketch.

mikhailcazi
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