Answer
The equation has 1 solution: $(1.755)$
Work Step by Step
$$x(x-1)^2=1$$ $$x(x^2-2x+1)=1$$ $$x^3-2x^2+x-1=0$$
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 $f(x)=x^3-2x^2+x-1$ as $x$ goes from $x_1$ to $x_2$.
- Take $x_1=1$: $f(1)=1^3-2\times1^2+1-1=1-2+1-1=-1\lt0$
- Take $x_2=2$: $f(2)=2^3-2\times2^2+2-1=8-8+2-1=1\gt0$
So there is a change of sign in $f(x)$ as $x$ goes from $1$ to $2$.
- On $[1,2]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(x^3-2x^2+x-1)=c^3-2c^2+c-1=f(c)$
So $f(x)$ is continuous on $[1,2]$.
Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[1,2]$ such that $f(c)=0$. In other words, the equation $x^3-2x^2+x-1=0$ or $x(x-1)^2=1$ has at least one solution in the interval $[1,2]$.
The graph of the function $f(x)=x^3-2x^2+x-1$ is enclosed below. Looking at the graph, it turns out that the curve $f(x)$ crosses the line $y=0$ at $1$ point with $x\in[1,2]$. The $x$-coordinates of that point is the solution of the equation.
So the equation has 1 solution: $(1.755)$
