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
\(f(x)\) 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
Show 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.
Updated On: Jun 16, 2026
- \(6,-6\)
- \(\sqrt3,-\sqrt3\)
- \(6\sqrt3,-6\sqrt3\)
- \(-6\sqrt3,6\sqrt3\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
A cubic polynomial with roots \(0,3,-3\) can be written as
\[
f(x)=k\,x(x-3)(x+3)
\]
The extrema are obtained by solving
\[
f'(x)=0.
\]
Step 1: Determine the polynomial. \[ f(x)=k\,x(x^2-9) \] Using \(f(1)=-8\), \[\begin{aligned} k(1)(1-9) &=-8\\ -8k &=-8 \end{aligned}\] \[ k=1 \] Hence \[ f(x)=x^3-9x \]
Step 2: Find critical points. \[\begin{aligned} f'(x) &=3x^2-9\\ &=3(x^2-3) \end{aligned}\] \[ f'(x)=0 \] gives \[ x=\pm\sqrt3 \]
Step 3: Evaluate the function at these points. At \(x=-\sqrt3\), \[\begin{aligned} f(-\sqrt3) &=(-\sqrt3)^3-9(-\sqrt3)\\ &=-3\sqrt3+9\sqrt3\\ &=6\sqrt3 \end{aligned}\] At \(x=\sqrt3\), \[\begin{aligned} f(\sqrt3) &=(\sqrt3)^3-9\sqrt3\\ &=3\sqrt3-9\sqrt3\\ &=-6\sqrt3 \end{aligned}\]
Step 4: Identify maximum and minimum values. \[ \text{Maximum value}=6\sqrt3 \] \[ \text{Minimum value}=-6\sqrt3 \] \[\begin{aligned} \boxed{ 6\sqrt3,\; -6\sqrt3 } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.
Step 1: Determine the polynomial. \[ f(x)=k\,x(x^2-9) \] Using \(f(1)=-8\), \[\begin{aligned} k(1)(1-9) &=-8\\ -8k &=-8 \end{aligned}\] \[ k=1 \] Hence \[ f(x)=x^3-9x \]
Step 2: Find critical points. \[\begin{aligned} f'(x) &=3x^2-9\\ &=3(x^2-3) \end{aligned}\] \[ f'(x)=0 \] gives \[ x=\pm\sqrt3 \]
Step 3: Evaluate the function at these points. At \(x=-\sqrt3\), \[\begin{aligned} f(-\sqrt3) &=(-\sqrt3)^3-9(-\sqrt3)\\ &=-3\sqrt3+9\sqrt3\\ &=6\sqrt3 \end{aligned}\] At \(x=\sqrt3\), \[\begin{aligned} f(\sqrt3) &=(\sqrt3)^3-9\sqrt3\\ &=3\sqrt3-9\sqrt3\\ &=-6\sqrt3 \end{aligned}\]
Step 4: Identify maximum and minimum values. \[ \text{Maximum value}=6\sqrt3 \] \[ \text{Minimum value}=-6\sqrt3 \] \[\begin{aligned} \boxed{ 6\sqrt3,\; -6\sqrt3 } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.
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