Animated Solution for Mathematics - Differentiation: Let $-1\le p\le 1$. Show that the equation $4x^3-3x-p=0$ has a unique root in the interval $[1/2,1]$ and identify it.

We are tasked with demonstrating that the equation $4x^3-3x-p=0$ possesses a unique root within the interval $[1/2,1]$ for the parameter range $-1\le p\le 1$.

Let us define the function $f(x)=4x^3-3x-p$. To understand the behavior of this function, we examine its rate of change.

We calculate the derivative:

Consider the interval of interest, $x\in [1/2,1]$. For any $x$ in this range, $x\ge 1/2$, which implies $x^2\ge 1/4$.

Multiplying by $12$, we obtain $12x^2\ge 3$. Therefore, the derivative satisfies:

This is a crucial realization. Because the derivative is non-negative, our function is strictly increasing, representing a steady, relentless climb.

Next, we apply the Intermediate Value Theorem (IVT). Let us evaluate the function at the boundaries of the interval.

At $x=1/2$: Since $p\ge -1$, it follows that $f(1/2)\le 0$.

At $x=1$: Since $p\le 1$, it follows that $f(1)\ge 0$.

Because the function is continuous and changes sign (or hits zero) between $1/2$ and $1$, there must be at least one root. Because the function is strictly increasing, that root is guaranteed to be unique.

To identify this root explicitly, we look at the expression $4x^3-3x$. This is the classic signature of the triple angle identity for cosine:

This identity is our key. We make the substitution $x=\cos (\theta )$. Our equation $4x^3-3x-p=0$ transforms into:

By the identity, this simplifies beautifully to:

We are now in the home stretch. To isolate $\theta$, we take the inverse cosine:

Finally, we recall our substitution $x=\cos (\theta )$. Substituting our value for $\theta$, we arrive at the exact, unique root:

We have successfully tamed a cubic polynomial by recognizing its hidden trigonometric structure. The unique root in the interval $[1/2,1]$ is given by $x=\cos \left(\frac{1}{3}{\cos }^{-1}p\right)$.