Understanding the Problem: Polynomial Functions and Functional Equations

We are given a polynomial function, let's call it \(f(x)\), and a condition it satisfies: when we apply the function twice, we get \(x^4\). This is written as \(f(f(x)) = x^4\). Our goal is to find the value of the derivative of \(f(x)\) at \(x=1\), which is denoted as \(f'(1)\).

Finding the Degree of the Polynomial

Let's first determine the degree of the polynomial \(f(x)\). Suppose the degree of \(f(x)\) is \(n\). When we compose a polynomial with itself, the degree of \(f(f(x))\) is the product of the degrees, i.e., \(n \times n = n^2\).

We are given that the degree of \(f(f(x))\) is the degree of \(x^4\), which is 4.

So, we have the equation for the degree:

\(n^2 = 4\)

Since \(f(x)\) is a polynomial, its degree \(n\) must be a non-negative integer. The possible integer solutions for \(n\) are \(n=2\) (since \(2^2=4\) and degree must be non-negative). A degree cannot be negative. Therefore, \(f(x)\) must be a polynomial of degree 2.

Identifying the Polynomial Function f(x)

Since \(f(x)\) is a polynomial of degree 2, it can be written in the general form:

\(f(x) = ax^2 + bx + c\)

where \(a, b, c\) are coefficients, and \(a \neq 0\).

Now, let's find \(f(f(x))\) by substituting \(f(x)\) into itself:

\(f(f(x)) = f(ax^2 + bx + c)\)

\(f(f(x)) = a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c\)

We know that \(f(f(x)) = x^4\). Let's expand the right side and compare coefficients with \(x^4\).

The term with the highest power of \(x\) in \(a(ax^2 + bx + c)^2\) comes from \(a(ax^2)^2 = a(a^2 x^4) = a^3 x^4\).

So, \(f(f(x)) = a^3 x^4 + (\text{lower degree terms})\).

Comparing the coefficient of \(x^4\) in \(f(f(x))\) with that in \(x^4\) (which is 1), we get:

\(a^3 = 1\)

Assuming real coefficients for the polynomial, the only real solution is \(a=1\).

So, \(f(x) = x^2 + bx + c\).

Let's substitute \(a=1\) back into the full expansion of \(f(f(x))\):

\(f(f(x)) = 1 \cdot (x^2 + bx + c)^2 + b(x^2 + bx + c) + c\)

\(f(f(x)) = (x^2 + bx + c)(x^2 + bx + c) + bx^2 + b^2x + bc + c\)

\(f(f(x)) = x^2(x^2 + bx + c) + bx(x^2 + bx + c) + c(x^2 + bx + c) + bx^2 + b^2x + bc + c\)

\(f(f(x)) = x^4 + bx^3 + cx^2 + bx^3 + b^2x^2 + bcx + cx^2 + bcx + c^2 + bx^2 + b^2x + bc + c\)

Now, group the terms by powers of \(x\):

\(f(f(x)) = x^4 + (b+b)x^3 + (c+b^2+c+b)x^2 + (bc+bc+b^2)x + (c^2+bc+c)\)

\(f(f(x)) = x^4 + 2bx^3 + (b^2 + 2c + b)x^2 + (2bc + b^2)x + (c^2 + bc + c)\)

We know this must be equal to \(x^4\). This means the coefficients of \(x^3, x^2, x^1\), and the constant term must all be zero.

• Coefficient of \(x^3\): \(2b = 0 \implies b = 0\).
• Coefficient of \(x^2\): \(b^2 + 2c + b = 0\). Substitute \(b=0\): \(0^2 + 2c + 0 = 0 \implies 2c = 0 \implies c = 0\).
• Coefficient of \(x^1\): \(2bc + b^2 = 0\). Substitute \(b=0, c=0\): \(2(0)(0) + 0^2 = 0\). This is consistent.
• Constant term: \(c^2 + bc + c = 0\). Substitute \(b=0, c=0\): \(0^2 + (0)(0) + 0 = 0\). This is consistent.

The only coefficients that satisfy the condition are \(a=1, b=0, c=0\). Therefore, the polynomial function is:

\(f(x) = 1x^2 + 0x + 0 = x^2\)

Let's quickly verify: \(f(f(x)) = f(x^2) = (x^2)^2 = x^4\). This confirms that \(f(x) = x^2\) is the correct polynomial function.

Calculating the Derivative at x=1

Now that we have found \(f(x)\), we need to find its derivative \(f'(x)\).

\(f(x) = x^2\)

Using the power rule for differentiation, \(\frac{d}{dx}(x^n) = nx^{n-1}\):

\(f'(x) = \frac{d}{dx}(x^2) = 2x^{2-1} = 2x\)

Finally, we need to evaluate \(f'(1)\) by substituting \(x=1\) into \(f'(x)\):

\(f'(1) = 2(1) = 2\)

Alternative Approach Using the Chain Rule

We are given \(f(f(x)) = x^4\). We can differentiate both sides with respect to \(x\).

Using the chain rule on the left side, \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Here, \(g(x) = f(x)\).

\(\frac{d}{dx}[f(f(x))] = f'(f(x)) \cdot f'(x)\)

The derivative of the right side is:

\(\frac{d}{dx}[x^4] = 4x^3\)

Equating the derivatives:

\(f'(f(x)) \cdot f'(x) = 4x^3\)

Now, evaluate this equation at \(x=1\):

\(f'(f(1)) \cdot f'(1) = 4(1)^3\)

\(f'(f(1)) \cdot f'(1) = 4\)

We need to find the value of \(f(1)\). Let's use the original equation \(f(f(x)) = x^4\) and substitute \(x=1\):

\(f(f(1)) = 1^4\)

\(f(f(1)) = 1\)

Let \(y = f(1)\). Then the equation becomes \(f(y) = 1\).

Since we determined that for a polynomial function satisfying \(f(f(x))=x^4\), the unique solution is \(f(x)=x^2\), let's use this. For \(f(x)=x^2\), \(f(1) = 1^2 = 1\). So, \(y=1\).

Substitute \(f(1)=1\) back into the differentiated equation \(f'(f(1)) \cdot f'(1) = 4\):

\(f'(1) \cdot f'(1) = 4\)

\((f'(1))^2 = 4\)

Taking the square root of both sides gives:

\(f'(1) = \pm 2\)

Since the polynomial function is \(f(x)=x^2\), its derivative is \(f'(x)=2x\). Evaluating at \(x=1\) gives \(f'(1)=2(1)=2\). This confirms that \(f'(1)=2\).

The value of \(f'(1)\) is 2.

Conclusion on f'(1)

Based on our analysis of the polynomial function \(f(x)\) satisfying the given condition, we found that \(f(x) = x^2\) is the unique polynomial solution with real coefficients. Calculating the derivative of this function at \(x=1\) gives \(f'(1) = 2\).

Revision Table: Key Concepts

Additional Information: Exploring Related Concepts

While we found \(f(x)=x^2\) as the polynomial solution with real coefficients, functional equations can sometimes have other types of solutions (e.g., non-polynomial functions). However, the question specifically asks for a polynomial function.

The step where we used \(f(1)=1\) from \(f(f(1))=1\) is crucial. For the specific polynomial solution \(f(x)=x^2\), \(f(1)=1^2=1\), so \(f(f(1))=f(1)=1\), which is consistent. If there hypothetically existed another polynomial solution, we might need to evaluate \(f(1)\) differently. However, our coefficient analysis strongly suggests \(f(x)=x^2\) is the only polynomial solution with real coefficients. If complex coefficients or non-polynomial functions were allowed, the analysis would be more complex and potentially yield other solutions for \(f(x)\) and possibly other values for \(f'(1)\).

The fact that \((f'(1))^2=4\) gave two possibilities, \(f'(1)=2\) and \(f'(1)=-2\), but only \(f'(1)=2\) arose from the confirmed polynomial solution \(f(x)=x^2\), and 2 is present in the options, supports 2 being the intended unique answer for a polynomial function.