3

Question:

If $6(8a + c) = 16b+ 3d$ then $f(x) = ax^3 + bx^2 + cx + d$ has at least one root in:

  1. $(-3,0)$
  2. $(-4,0)$
  3. $(-4,-3)$
  4. $(0,2)$

Attempt: Having solved several such questions, there is usually a hint hidden somewhere within the question. However, I can't find any hint anywhere in this question. I don't see any application of Rolle's theorem. I know for a fact that I could use the property that if $f(x)$ has a root in $(a,b)$ then $f(a)f(b) < 0$, but don't see any way of doing so unless I substitute every option into it. Any hint would be appreciated.

Gummy bears
  • 3,408
  • @Casteels Aha..... I posted the question while looking at my solution, which involved an attempt at differentiation, and thus the quadratic equation. Just a moment...... And yeah thanks for that. Stupid of me. – Gummy bears Jul 14 '15 at 15:41
  • It seems unclear to me what you mean with these roots, I assume you are trying to find the values of $x$ where $f(x)=0$. Why are there 2 numbers in that case? Your function is defined as $f: \mathbb{R} \rightarrow \mathbb{R}$ right? – Jan Jul 14 '15 at 15:58
  • @Jan Those are the ranges in which the roots may be present. Basically, it means that the value of $x$ such that $f(X) = 0$ may be in between the two numbers given. – Gummy bears Jul 14 '15 at 16:00
  • Of course, it is really obvious, I assumed they were points and not intervals, my mistake. – Jan Jul 14 '15 at 16:01
  • From the first equation we see $c=-8a$ and $d=\frac{-16}{3} c$. Substituting this into $f$ gives us: $$f(x)=ax(x^2-8a) + b(x^2-16/3)$$ This should help you a bit. – Jan Jul 14 '15 at 16:05
  • @Jan Slight typo in the question. No equality with zero. All fixed now. Sorry :( – Gummy bears Jul 14 '15 at 16:12

1 Answers1

1

We are given $48a-16b+6c-3d=0 \tag{1}$.

Rolle's theorem states that if a real-values function has equal values at two points then it must have a local extrema between them. So we need to find the anti-derivative $F(x)$ of $f(x)$ in order to apply:

$F(x) = \frac{1}{4}ax^4 + \frac{1}{3}bx^3 + \frac{1}{2}cx^2 + dx + k \tag{2}$

Then

$F(0) = k$.

Also,

$F(-4) = 64a - \frac{64}{3}b + 8c - 4d + k = \dfrac{4}{3}(48a-16b+6c-3d) + k = 0 = k = k$ by (1).

Hence $F(0) = F(-4)$ so $F(x)$ has a local extrema in $(-4,0)$ and so $f(x)$ has a root in $(-4,0)$.

Marconius
  • 5,635