# Chapter 2 - Section 2.5 - Continuity - Exercises - Page 96: 76

The equation has 3 solutions: $(-3.906, 0.067, 3.839)$. $$x^3-15x+1=0$$ 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 $f(x)=x^3-15x+1$ as $x$ goes from $x_1$ to $x_2$. - Take $x_1=0$: $f(0)=0^3-15\times0+1=1\gt0$ - Take $x_2=2$: $f(2)=2^3-15\times2+1=8-30+1=-21\lt0$ So there is a change of sign in $f(x)$ as $x$ goes from $0$ to $2$. - On $[0,2]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(x^3-15x+1)=c^3-15c+1=f(c)$ So $f(x)$ is continuous on $[0,2]$. Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[0,2]$ such that $f(c)=0$. In other words, the equation $x^3-15x+1=0$ has at least one solution in the interval $[0,2]$. The graph of the function $f(x)=x^3-15x+1$ is enclosed below. Looking at the graph, it turns out that the curve $f(x)$ crosses the line $y=0$ at $3$ points. The $x$-coordinates of these points are the solutions of the equation. So the equation has 3 solutions: $(-3.906, 0.067, 3.839)$. 