Question:
If $6(8a + c) = 16b+ 3d$ then $f(x) = ax^3 + bx^2 + cx + d$ has at least one root in:
- $(-3,0)$
- $(-4,0)$
- $(-4,-3)$
- $(0,2)$
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.
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 $(-4,0)$.