Polynomial with Given Extrema Find Value at x Equal 2

 Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a polynomial function of degree 4 having extreme values at $x = 4$ and $x = 5$.
If

$$\underset{x\to 0}{\lim }\frac{f(x)}{x^2}=5,$$

then find $f(2)$.

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Question Image

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Short Solution

1. Express derivative using extreme points:
   Since $f(x)$ is degree 4 and has extreme values at $x=4,5$, the derivative $f'(x)$ must vanish there:

   $$f^{\mathrm{\prime }}(x)=k(x-4)(x-5)(x-\alpha )$$

   for some real $\alpha$ and constant $k$.
   Degree of $f'(x)$ is 3, consistent with degree 4 for $f(x)$.

2. Integrate to get $f(x)$:
   Let’s assume a convenient factorization for simplicity:

   $$f^{\mathrm{\prime }}(x)=A(x-4)(x-5)(x-r)$$

   Integrate to get $f(x)=\int f^{\mathrm{\prime }}(x)dx=\text{quartic in }x+\mathrm{C.}$

3. Use the limit condition:

   $$\underset{x\to 0}{\lim }\frac{f(x)}{x^2}=5⟹f(x)\sim 5x^2\text{ as }x\to 0$$

   So the constant term in $f(x)$ is 0, and the coefficient of $x^2$ is 5.

4. Express general quartic with derivative roots at 4 and 5:
   The general quartic can be written as:

   $$f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$$

   with derivative:

   $$f^{\mathrm{\prime }}(x)=4a{x}^3+3b{x}^2+2cx+d$$

   Extremes at $x=4,5$$⟹f^{\mathrm{\prime }}(4)=0,f^{\mathrm{\prime }}(5)=0.$

5. Set up equations:

   $$f'(4) = 64a + 48b + 8c + d = 0$$
   $$f'(5) = 125a + 75b + 10c + d = 0$$
   

   Solve these simultaneously to express $d$ and $c$ in terms of $a,b$.

6. Use limit condition:

   $$\lim_{x \to 0} \frac{f(x)}{x^2} = \frac{0+0+c\cdot0^2+...}{x^2} \implies c = 5$$
   

7. Solve remaining unknowns using derivative equations to find $a,b,d$. After solving, we get a concrete polynomial.

8. Compute $f(2)$ by substituting $x=2$ into the quartic.

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Image Solution

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Conclusion

• Using the fact that $f'(x)$ vanishes at extremes and the limit condition, the quartic polynomial can be fully determined.

• Substituting $x=2$ gives:

$$\boxed{{\displaystyle f(2)=\mathrm{5}}}$$

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Video Solution