If $P(x)=2013x^{2012}-2012x^{2011}-16x+8$, then $P(x)=0$ for $x\in\left[0,8^{\frac{1}{2011}}\right]$ has
- exactly one real root.
- no real root.
- at least one and at most two real roots.
- at least two real roots.
This question has answers here and here.
I have doubts in them. I'll list that below.
Answer$1:$
Let $Q(x) = x^{2013} – x^{2012} -8x^2 +8x + c $. Note that $Q'(x)=P(x)$. Since $Q(0)=Q(1)=Q(8^\frac{1}{2011})$, we apply Rolle's Theorem to each of the intervals $[0,1]$ and $[1,8^\frac{1}{2011}]$ to conclude that $Q'(x)=P(x)$ has at least a root in each of the intervals $(0,1)$ and $(1,8^\frac{1}{2011})$. Hence $P(x)$ has at least two roots.
Doubt: How do we know $1\in(0,8^\frac1{2011})?$
Answer$2:$
$$\begin{align}P(x)&=2013x^{2012}-2012x^{2011}-16x+8 \\
P'(x)&=2013\cdot 2012x^{2011}-2012\cdot 2011x^{2010}-16\end{align}$$
$P'(0)$ is negative and $P'\left(8^{\frac{1}{2011}}\right)$ is positive. So $P'(x)$ has at least one real root.$\implies P(x)$ has at least two real roots.
Doubt: How do we know $P'(8^\frac1{2011})\gt0?$
Answer$3:$
We have $p(0)=8,p(1)=-7$ so there is at least one root in [0,1]. Put $\alpha=8^{1/2011}$. Then $p(\alpha)$ is approx 0. So we may have another root just inside the interval $[0,\alpha]$.
Doubt: How do we know $P(\alpha)$ is close to zero?
As I am typing this, I am realizing all my doubts are just regarding calculating the value at $x=8^\frac1{2011}$. I am not able to convincingly tell its value or sign, without using calculator.
Best Answer
$x=8^{\frac {1}{2011}}$ means that $x^{2011}=8$.
If $x≤1$, then its 2011th power would only be smaller than or equal to 1, so we must have $1<x$.