Question

\(f(x)\) is a polynomial of degree \(3\) such that \[ f(0)=f(3)=f(-3)=0 \] \[ f(1)=-8 \] The maximum and minimum values of \(f(x)\) are respectively

• A. \(6,-6\)
• B. \(\sqrt3,-\sqrt3\)
• C. \(6\sqrt3,-6\sqrt3\)
• D. \(-6\sqrt3,6\sqrt3\)

Hint:

If a cubic polynomial has roots \(a,b,c\), write \[ f(x)=k(x-a)(x-b)(x-c) \] first, determine \(k\) using the given condition, then use \[ f'(x)=0 \] to find maxima and minima.

Answer 1

The Correct Option is C

Step 1: Build the cubic from its roots.
Since $f(0) = f(3) = f(-3) = 0$, the roots are $0, 3, -3$. So $f(x) = k\,x(x-3)(x+3) = k\,x(x^2 - 9)$.

Step 2: Use the extra condition to find $k$.
Given $f(1) = -8$: $f(1) = k(1)(1 - 9) = -8k$. Setting $-8k = -8$ gives $k = 1$.

Step 3: Write the actual function.
So $f(x) = x^3 - 9x$.

Step 4: Find the turning points.
$f'(x) = 3x^2 - 9 = 3(x^2 - 3)$. Setting it to $0$ gives $x = \sqrt 3$ and $x = -\sqrt 3$.

Step 5: Evaluate the function at those points.
At $x = -\sqrt 3$: $f = (-\sqrt 3)^3 - 9(-\sqrt 3) = -3\sqrt 3 + 9\sqrt 3 = 6\sqrt 3$. At $x = \sqrt 3$: $f = 3\sqrt 3 - 9\sqrt 3 = -6\sqrt 3$.

Step 6: Identify max and min.
The larger value $6\sqrt 3$ is the maximum, and $-6\sqrt 3$ is the minimum. \[ \boxed{6\sqrt 3,\ -6\sqrt 3} \]