Find the number of real solutions to the equation $ x^7=x+1$.
The equation $x^7 = x+1 $ has:
a) no real solution
b) no positive real solution
c) a real solution in the interval (0,2)
d) a real solution but not within (0,2)
Which is the correct answer and why ? How do i find the answer to such questions?
$\endgroup$ 15 Answers
$\begingroup$Consider the 2 equations:
$$y = x^7 \tag{A}$$ $$y = x + 1 \tag{B}$$
What does equation (A) look like? It has an intersection point at $(0, 0)$, grows incredibly fast before $x=-1$ and after $x=1$.
Equation (B) grows so much slower that it can only intersection (A) at one point. Since $1^7 < 1 + 1 < 2^7$, the intersection point is somewhere between $x=1$ and $x=2$. So the answer is $c$.
$\endgroup$ $\begingroup$Use Descartes' Rule of Signs on $f(x)=x^7-x-1$:
It states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.
So we have one positive root. Now evaluate $f(x)$ at $0$ and $2$ to judge whether the root lies inside or outside of this interval...
$\endgroup$ $\begingroup$The critical points of $f(x) = x^7 - x - 1$ are given by $f^{\prime}(x) = 0 \ \Rightarrow \ x = \pm \sqrt[6]{\frac{1}{7}} \simeq \pm 0.723$. Note that $f^{\prime \prime}(x) = 42x^5$. On the order hand $f(\pm 0.723) < 0$ and being $\displaystyle{\lim_{x \to -\infty}}f(x) = - \infty$ and $\displaystyle{\lim_{x \to +\infty}}f(x) = +\infty$. Therefore, $f(x) = 0$ has only one real solution.
$\endgroup$ $\begingroup$To solve a question like this, let's use Descartes's rule of signs.
So our function is $f(x) = x^7 - x - 1 $
We then do the following:
$f(-x) = -x^7 + x - 1$
There is one change in sign for f(x), meaning there is one positive real root.
There are two changes for f(-x), meaning there are either 2 or 0 negative real roots.
Since there are 7 roots, by the fundamental theorem by algebra, we could have 1 positive root, 2 negative roots, and 4 complex roots.
Or we could have 1 positive root, 0 negative roots, and 6 complex roots.
We know that the answer is either c or d.
So, we can check values.
$f(0) = -1$, and $f(2) = 2^7 - 2 -1 = 125$
Since from $(0,2)$ we have $f(x)$ change from negative and positive. Therefore, by the IVT (Intermediate Value theorem), there is at least one zero in that open interval.
The answer is c.
$\endgroup$ $\begingroup$You could also not use Descartes' rule of signs (I didn't even hear about it until now. Weird.) You want to find the roots of the polynomial $x^7 - x - 1$. Knowing the available answers, you can evaluate at $0$, which yields $-1$, and at $2$, which yields $125$. Then by the intermediate value theorem, there is a root in the interval $(0,2)$. The only possible answer is then c).
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