Answer
Take the function $f(x)=x^3-15x+1$ and prove that $f$ has crossed the line $y=0$ three times in the interval $[-4,4]$, using the Intermediate Value Theorem.
Work Step by Step
$$x^3-15x+1=0$$
To prove the equation has 3 solutions in the interval $[-4,4]$, we can do the followings:
- Take the function $f(x)=x^3-15 x+1$
- Show that $f(x)$ is continuous on $[-4,4]$.
- Prove the $f$ has changed its sign (negative to positive and vice versa) 3 times in the interval $[-4,4]$, meaning each time $f$ must cross the line $y=0$ once according to Intermediate Value Theorem.
Considering the function $f(x)=x^3-15 x+1$:
- Domain: $(-\infty,\infty)$
1) Continuity:
On $(-\infty,\infty)$, we notice that $\lim_{x\to c}(x^3-15 x+1)=c^3-15c+1=f(c)$
That means $f$ is continuous on $(-\infty,\infty)$, so $f$ is continuous on $[-4,4]$ as well.
2) Show that $f$ has changed its sign 3 times in $[-4,4]$
Here I graph $f(x)=x^3-15 x+1$ for better findings, which I have also enclosed below.
- At $x=-4$: $f(-4)=(-4)^3-15(-4)+1=-64+60+1=-3\lt0$
- At $x=0$: $f(0)=0^3-15(0)+1=1\gt0$
- At $x=2$: $f(2)=2^3-15(2)+1=8-30+1=-21\lt0$
- At $x=4$: $f(4)=4^3-15(4)+1=64-60+1=5\gt0$
Here, according to the Intermediate Value Theorem, because $f$ is continuous on $[-4,4]$, in the interval $[-4,0]$, there exists a value of $x=c$ such that $f(c)=0$, or $f$ has crossed $y=0$ once here.
Similarly, each in the interval $[0,2]$ and $[2,4]$, $f$ must cross $y=0$ once there.
In other words, the equation $x^3-15x+1=0$ has 3 solutions in the interval $[-4,4]$.
