# How can I solve for the intersection points of y=(x^2)(e^-x) and y=1-2sinx over 2<=x<=7 ?

## I know that there should be 2 intersections within this domain, but which would otherwise be cyclic over $\mathbb{R}$ (from what I saw graphically).

Aug 12, 2017

I know of no way to solve this, using algebraic methods. I recommend the use of WolframAlpha

#### Explanation:

Given:

$y = {x}^{2} {e}^{-} x \text{ [1]}$
$y = 1 - 2 \sin \left(x\right) \text{ [2]}$
$2 \le x \le 7 \text{ [3]}$

If we attempt to find x coordinates of the points of intersection by asserting $y = y$, we obtain the the equation:

x^2e^-x =1 - 2sin(x); 2 <= x <=7

I know of no way to solve this, using algebraic methods. One can use numerical analysis methods such as Newton's Method to find the approximate x values and, then use one of the equations, to obtain the corresponding values of y.

But it is easier to return to the two equations and the domain restriction and give them to WolframAlpha , because the computation engine will give you x and y at the same time.

x≈2.87172, y≈0.46678

and

x≈6.77681, y≈0.0523502

Aug 12, 2017

See below.

#### Explanation:

Calling $f \left(x\right) = {x}^{2} {e}^{-} x + 2 S \in x - 1$ we are asking for the solutions for $f \left(x\right) = 0$ such that $x \in \left[2 , 7\right]$

The iterative process to determine the solutions to

$f \left(x\right) = 0$

is inspired in the linear approximation using the Taylor expansion of $f \left(x\right)$ near the solution ${x}^{\circ}$ Using the Taylor expansion near ${x}^{\circ}$ we have

$f \left({x}_{k + 1}\right) \approx f \left({x}_{k}\right) + f ' \left({x}_{k}\right) \left({x}_{k + 1} - {x}_{k}\right) + O \left({\left\mid {x}_{k + 1} - {x}_{k} \right\mid}^{2}\right)$

Now if ${x}_{k + 1}$ is near ${x}^{\circ}$ then $f \left({x}_{k + 1}\right) \approx 0$ and also $O \left({\left\mid {x}_{k + 1} - {x}_{k} \right\mid}^{2}\right) \approx 0$ so we follow with

$0 = f \left({x}_{k}\right) + f ' \left({x}_{k}\right) \left({x}_{k + 1} - {x}_{k}\right)$ or

${x}_{k + 1} = {x}_{k} - f \frac{{x}_{k}}{f ' \left({x}_{k}\right)}$ which is the celebrated formula attributed to Newton.

Now beginning with a guess near ${x}^{\circ}$ we have in many cases a convergent sequence like

((x_k, f(x_k)),(2.6, 0.533092), (2.89136, -0.0407656), (2.87177, -0.000110522), (2.87172, -8.60538*10^-10), (2.87172, -4.44089*10^-16))

or

((x_k, f(x_k)),(5.5, -2.28746),(7.20875, 0.636384),(6.66714, -0.194282),(6.77419, -0.00452637),(6.77681, \ -3.1636*10^-6),(6.77681, -1.55442*10^-12))

So we found two roots

$2.871718639488265$ and $6.776814476412012$