Question

Question #2096b

Answer 1

Three solutions are (to 10dp):

`x_1 = -0.3714177525`
`x_2 = 0.6052671213`
`x_3 = 4.7079379181`

Explanation:

Let:

`f(x) = e^x-5x^2`

Our aim is to solve `f(x)=0`. First let us look at the graph:
graph{e^x-5x^2 [-5, 10, -30, 10]}

We can see that there are three solutions; one solution in the interval `-1 lt x lt 0`, one in `0 lt x lt 1`, and one in `4 lt x lt 5`. The root we find will depend upon our initial approximation `x_0`, and this value will require a little trial and error.

To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

`{ (x_1,=x_0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}`

Therefore we need the derivative:

`\ \ \ \ \ \ \f(x) = e^x-5x^2`
`:. f'(x) = e^x-10x`

Then using excel working to 10dp we can tabulate the iterations as follows:

Initial Value `x_0=-1`

Initial Value `x_0=1`

Initial Value `x_0=5`

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the three solutions are (to 10dp):

`x_1 = -0.3714177525`
`x_2 = 0.6052671213`
`x_3 = 4.7079379181`