Show that the equation $x^3 +1 = 15x$ has three solutions in the interval $[-4,4]$ Show that the equation $x^3 +1 = 15x$ has three solutions in the interval $[-4,4]$
There are many elementary ways to solve this. But this question came out in exam for calculus!!
How do i solve it using calculus?
Here's what I learnt so far... Continuity, derivatives, mean value theorem, rolles theorem, fermats theorem, limits.
 A: Continuity should do it. Evaluate $x^3-15x+1$ at enough places between $-4$ and $4$ to show that it changes sign three times. 
A: Let $f(x)=x^3-15x+1$; then $f\,'(x)=3x^2-15=3(x^2-5)=3(x-\sqrt5)(x+\sqrt5$. It’s easy to check that $f$ has a local maximum at $x=-\sqrt5$ and a local minimum at $x=\sqrt5$, either by the second derivative test, by examining the sign of the first derivative, or simply by knowing the shape of the graph of a cubic polynomial. 
Now 
$$\begin{align*}
f(-4)&=-64+60+1=-3<0,\\
f(-\sqrt5)&=-5\sqrt5+15\sqrt5+1=10\sqrt5+1>0,\\
f(\sqrt5)&=5\sqrt5-15\sqrt5+1=-10\sqrt5+1<0,\text{ and}\\
f(4)&=64-60+1=5>0\;,
\end{align*}$$
so $f(x)$ changes sign three times on the interval $[-4,4]$ and by the intermediate value theorem must have (at least) three zeroes.
Note that there are many ways to choose the test points, but using the critical points is an easy way to be sure of hitting good ones.
Added: To show that there are only three solutions to the original equation, apply Rolle’s theorem to $f$ to show that between any two zeroes of $f$ there must be a critical point.
A: You could use the intermediate value theorem. Write $f(x)=x^3 -15x+1$ and find pairs of positive and negative points on the interval. e.g. $f(0)=1$ and $f(1)=-13$. Use the intermediate value theorem to show that, since the curve is continuous, it must cross the x-axis. Continuing from the previous example, there exists an $x$ in $[0,1]$ s.t. $f(x)=0$.
