Where does the root of this equation lie$?$

Question

If $0\le p\le16$, then the equation $x^3-12x-p=0$ has one root in

$(2,3)$

$(3,4]$

$(4,5)$

None of these

My work:

I know that the function $f(x)=x^3-12x-p=0$ is strictly increasing in the intervals $(-\infty,-2)$ and $(2,\infty)$.

I also know that if $f(a)$ and $f(b)$ are of opposite signs then at least $1$ or in general odd number of roots of the equation $f(x)=0$ lie between $a$ and $b$.

But I can't use these two pieces of information here.

Any help is greatly appreciated.

Answer 1

It can be helpful to start by "discarding" the constant term: $ \ x^3 - 12x \ $ has zeroes $ \ 0 \ , \ \pm 2 \sqrt3 \ \approx \ \pm 3.5 \ \ $ and extrema at $ \ (-2 \ , \ 16) \ $ and $ \ (+2 \ , \ -16) \ \ . $ So shifting this function "downward" by $ \ 16 \ $ places the relative maximum on the $ x-$axis, making it a "double zero". A little synthetic/polynomial division then gives us $ \ x^3 - 12x - 16 \ = \ (x + 2)^2·(x - 4) \ \ . $ So the "downward shift" keeps one zero within $ \ (3 \ , \ 4] \ \ . $

Answer 2

Hint. You know the function is increasing from $x=2$ on so it has at most one root there. What are its values at $2$, $3$ and $4$?

Answer 3

Acrually I got the answer. Please check if I'm right.

$f(3)=-9-p$ which is always $\lt0$ and $f(4)=16-p$ which is either $\gt0$ or equal to $0$ or we can say that $f(3)\cdot f(4)\lt0$ or equal to $0$ $\implies$ there is one root in the interval $(3,4]$ (there is square bracket at the end because $f(4)$ can be $0$)