$x$ intercept problem

Question

How would I find the $x$ intercept of $x^5-x^3+2=0$? I haven’t learned about things like synthetic division or any theorems, just algebraic manipulations.

Answer 1

By the rational root theorem, if the function $f(x) = x^5 - x^3 + 2 = 0$ has a rational zero, then it must be among the rational numbers:

$1, -1, 2, -2$

However, we see that if we plug in any of these values of $x$ into our function, we will not get $0$. That is:

$f(1) = 2, \quad f(-1) = 2, \quad f(2) = 26, \quad f(-2) = -22$

Therefore, we may conclude that any value of $x$ which satisfies $x^5 - x^3 + 2 =0$ will not be a rational number (so, at best, we can only approximate the value of $x$ which does so). One of our options is to simply use a computer to approximate the solution for us. Otherwise, we would probably have to use Newton's Method to approximate this solution by hand (which is very tedious).

Answer 2

My answer is not totally correct but I hope it helps a little :

Consider $f(x)=x^5-x^3+2$ . we have $f'(x)=5x^4-3x^2$ and $f'(x)=0$ if $x=0$ or $x=\pm \sqrt{\frac{3}{5}}$ and it is also possible to determine the sign of $f'(x)$ for all other $x$.

Note that $5x^4-3x^2>0$ if $5x^4>3x^2$ if $x^2>3/5$ . that is, $x >+ \sqrt{\frac{3}{5}}$ or $x <- \sqrt{\frac{3}{5}}$. Also $5x^4-3x^2<0$ if $- \sqrt{\frac{3}{5}}<x<+ \sqrt{\frac{3}{5}}$.

Thus $f$ is increasing for $x <- \sqrt{\frac{3}{5}}$, decreasing between $- \sqrt{\frac{3}{5}}$ and $+ \sqrt{\frac{3}{5}}$ and again increasing for $x>+ \sqrt{\frac{3}{5}}$.

For this $f$, finding $x$ for $f(x)=0$ is difficult. But IVT guranteed that such a number is between $-2$ and $-1$, since $f(-2)<0$ and $f(-1)>0$

with this information the graph look like this:(with the help of Desmos)

Answer 3

$$f(x)=x^5-x^3+2\implies f'(x)=5x^4-3x^2\implies f''(x)=20x^3-6x$$

The first derivative cancels twice at $x=0$ and at $x=\pm \sqrt{\frac{3}{5}}$.

$$f(0)=+2 $$ $$f\left(+\sqrt{\frac{3}{5}} \right)=2-\frac{6 \sqrt{\frac{3}{5}}}{25}\approx 1.8141 \qquad f''\left(+\sqrt{\frac{3}{5}} \right)=6 \sqrt{\frac{3}{5}}$$ $$f\left(-\sqrt{\frac{3}{5}} \right)=2+\frac{6 \sqrt{\frac{3}{5}}}{25}\approx 2.1859 \qquad f''\left(-\sqrt{\frac{3}{5}} \right)=-6 \sqrt{\frac{3}{5}}$$ So, ther is only one real root which is negative. By inspection, $f(-1)=2$, $f(-2)=-22$; so the root is close to $-1$.

To approximate the root, use Taylor series to get $$f(x)=2+2 (x+1)-7 (x+1)^2+O\left((x+1)^3\right)$$ and the root to be considered is given by $$x=\frac{1}{7} \left(\sqrt{15}-6\right)\approx -1.41043$$ Now, use Newton method which will update the guess according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=\frac{-4 x_n^5+2 x_n^3+2}{3 x_n^2-5 x_n^4}$$ to get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & -1.410426192 \\ 1 & -1.354288685 \\ 2 & -1.347943175 \\ 3 & -1.347867906 \\ 4 & -1.347867896 \end{array} \right)$$ which is the solution for ten significant figures.