Drawing the graph of cubic function

Question

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: $[-2,2]$?

Answer 1

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 $[-2,2]$.

Answer 2

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.