Answer
The equation has 3 solutions: $(-0.855, 0.403, 1.452)$.
Work Step by Step
$$2x^3-2x^2-2x+1=0$$
To prove the equation has a solution using the Immediate Value Theorem, we need to find 2 values $x_1$ and $x_2$ and show that there is a change of sign in the function $f(x)=2x^3-2x^2-2x+1$ as $x$ goes from $x_1$ to $x_2$.
- Take $x_1=0$: $f(0)=2\times0^3-2\times0^2-2\times0+1=1\gt0$
- Take $x_2=1$: $f(1)=2\times1^3-2\times1^2-2\times1+1=2-2-2+1=-1\lt0$
So there is a change of sign in $f(x)$ as $x$ goes from $0$ to $1$.
- On $[0,1]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(2x^3-2x^2-2x+1)=2c^3-2c^2-2c+1=f(c)$
So $f(x)$ is continuous on $[0,1]$.
Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[0,1]$ such that $f(c)=0$. In other words, the equation $2x^3-2x^2-2x+1=0$ has at least one solution in the interval $[0,1]$.
The graph of the function $f(x)=2x^3-2x^2-2x+1$ is enclosed below. Looking at the graph, we find that the curve $f(x)$ crosses the line $y=0$ in fact 3 times. The $x$-coordinates of these points are the solutions of the equation.
So the equation has 3 solutions: $(-0.855, 0.403, 1.452)$
