Limits

Question

Let $f(x)$ be a polynomial of degree 4 having extreme values at $x=1$ and $x=2$. If $\lim_{x 	o 0} [1 + f(x)/x^2] = 3$, then $f(2)$ is equal to:

RankGuru · Accepted Answer

1. Analyze the limit:
$\lim_{x 	o 0} (1 + \frac{f(x)}{x^2}) = 3$ implies $\lim_{x 	o 0} \frac{f(x)}{x^2} = 2$.
For this limit to exist and be finite, $f(x)$ must have 0 coefficients for $x^0$ and $x^1$ terms, and the coefficient of $x^2$ must be 2. Since $f(x)$ is degree 4, let $f(x) = ax^4 + bx^3 + 2x^2$.

2. Use derivative conditions:
$f(x)$ has extremes at $x = 1$ and $x = 2$, so $f'(1) = 0$ and $f'(2) = 0$.
$f'(x) = 4ax^3 + 3bx^2 + 4x = x(4ax^2 + 3bx + 4)$.

At $x=1$: $4a + 3b + 4 = 0$  (Eq 1)
At $x=2$: $32a + 12b + 8 = 0 \implies 8a + 3b + 2 = 0$ (Eq 2)

3. Solve for a and b:
Subtract Eq 1 from Eq 2: $(8a + 3b + 2) - (4a + 3b + 4) = 0 \implies 4a - 2 = 0 \implies a = 1/2$.
From $4a + 3b = -4$: $2 + 3b = -4 \implies 3b = -6 \implies b = -2$.

So $f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$.

4. Calculate $f(2)$:
$f(2) = \frac{1}{2}(16) - 2(8) + 2(4) = 8 - 16 + 8 = 0$.

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