Numerical methods for solving nonlinear equations I have this equation 

$x^{5}+x^{3}+3=0$

and I'm supposed to find one root with a given accuracy using the secant method. I was wondering how I can localize a solution to an interval so I can apply the formula. I'm aware that I could separate the equation into two functions and plot both of them, but I think that can be hard sometimes when I don't have access to plotting tools like Desmos and my equation gets complicated.
 A: We take a few guesses for the function $f(x) = x^5+x^3+3$
First, we plug in $x=-1$. We find that the equation is positive, but for $x=-2$, the equation is negative.
So, we know to search in the range of $(-2,-1)$. We set $x_0=-2$ and $x_1=-1$.
Now, we use the formula for the secant method:
$$x_{i+1}=x_i-\frac{f(x_i)\cdot(x_i-x_{i-1})}{f(x_i)-f(x_{i-1})}$$
Applying this repeatedly gets us the sequence $$-2, -1, -1.0263157894736843, -1.119763138322457, -1.1034910955412918, -1.105259392177301, -1.105298653114797, -1.1052985459998306, -1.1052985460061695, -1.1052985460061695$$This is how we converge upon the solution.
A: As suggested, an IVT is something useful:
Note that $f(-2)<0<f(-1)$, where $f(x)=x^5+x^3+3$. Hence by the IVT you have a solution in $(-2,-1)$
A: Choose two points $x_0$and $x_1$
Here I choose $x_0=-1$ and $x_1=-2$
$$f(x_0)f(x_1)<0$$
Then apply the secant formula 
$$x_n=\frac{x_{n-2}f(x_{n-1})-x_{n-1}f(x_{n-2})}{f(x_{n-1}-f(x_{n-2}))}$$
Thus $$x_2=-1.02631$$
$$x_3=-1.0464$$
$$x_4=-1.1130$$
$$x_5=-1.1045790$$
$$x_6=-1.1052901$$
$$x_7=-1.1052986$$
$$x_8=-1.1052985$$
Keep continuing until you get the root.
