Find the second derivative for the function 5x^3+60x^2-36x-41
and solve the equation F(X)=0
i got to the second derivative but what do we do next to be able to get one of these answers..
F(x) = 5x3+60x2−36x−41
1. derivative of F(x)
with respect to x = 15x2+120x−36
2. derivative of F(x)
with respect to x = 30x+120
a. 4
b. 5
c. 0
d. 41
e. 36
Your derivatives are correct
As to solving f(x) = 0
5x^3+60x^2-36x-41 = 0 is a tough one, since it does not factor
I used an apps to find
x = 1.11169
x = -12.5227
x = -.589024
To solve the equation F(x) = 0, you need to find the values of x that satisfy this equation. This can be done by setting the function equal to zero and solving for x.
So, for the given function F(x) = 5x^3 + 60x^2 - 36x - 41, we can set it equal to zero:
5x^3 + 60x^2 - 36x - 41 = 0
Now, to solve this equation, there are different methods you can use, such as factoring, synthetic division, or using the quadratic formula if the equation can be manipulated into a quadratic form. In this case, factoring might be challenging, so let's use a numerical method called the Newton-Raphson method to approximate the solution.
Here's how you can apply the Newton-Raphson method to find the solution:
1. Start with an initial guess for x, let's say x0 = 1.
2. Use the formula: x1 = x0 - (F(x0) / F'(x0)), where F'(x) is the first derivative of F(x).
3. Calculate the first derivative, F'(x) = 15x^2 + 120x - 36.
4. Plug in the values x0 and F'(x0) into the formula to get the first approximation, x1.
x1 = x0 - (F(x0) / F'(x0))
x1 = 1 - ((5(1)^3 + 60(1)^2 - 36(1) - 41) / (15(1)^2 + 120(1) - 36))
x1 = 1 - (5 + 60 - 36 - 41) / (15 + 120 - 36)
x1 = 1 - (88 / 99)
x1 = 1 - 0.888888...
x1 ≈ 0.111111...
Repeat steps 2-4 with the new approximation (x1) until you reach a desired level of accuracy or until the process converges to a solution.
By continuing the steps with the updated x value, you can find a better approximation for the solution.