For the function y=(e^2x)+(3x)-(10), use Newton's method and the calculator method to find the x value for which y=15. Please show your work so that I can understand the question! Thank you so much!! it means alot!!
so you are solving
e^(2x) + 3x - 25 = 0
let y = e^(2x) + 3x - 25
y' = 2e^(2x) + 3
newtons's formula says
X = x - f(x)/f'(x)
where x is your starting x value and X is the new value.
Hopefully your X will approach x, when that happens you have the solution.
X = x - (e^(2x) + 3x - 25)/(2e^(2x) + 3)
= (2xe^(2x) - e^(2x) + 25)/(2e^2(2x) + 3)
I made a rough sketch of y = e^(2x) + 3x - 25
and noticed that there was an x-intercept, which would be your solution, at appr x = 1.5
So I will make that my starting x
X ---- x
-------- 1.5
1.5096 ---1.50915695
1.50951695 -- 1.509515688
1.509515688 -- 1.50951688
Wow, got the answer correct to 9 decimal places after only 3 iterations.
The key thing is that you start with an initial guess close to the real number.
To find the x value for which y = 15 in the given function y = e^(2x) + 3x - 10, we can use Newton's method and the calculator method. Let's go step by step:
1. Newton's Method:
a) Start with an initial guess for the x value, let's say x0.
b) Calculate the derivative of the function, y' = 2e^(2x) + 3.
c) Use the following iteration formula to update the guess:
x1 = x0 - (f(x0) - 15) / f'(x0),
where f(x0) is the value of the function at x0, and f'(x0) is the value of the derivative at x0.
d) Repeat step c) until the guess converges to the desired accuracy. In this case, we want to find the x value for which y = 15.
2. Calculator Method:
a) Enter the function y = e^(2x) + 3x - 10 into a graphing calculator.
b) Graph the function and find the point where the graph intersects the line y = 15.
You can visually locate the point or use the intersection feature on the calculator.
c) Read the x-coordinate of the intersection point, which will give you the x value we are looking for.
Now, let's perform the calculations using Newton's method. For simplicity, let's assume an initial guess of x0 = 0.
1a) Start with x0 = 0.
1b) Calculate the derivative: y' = 2e^(2x) + 3.
So, y' = 2e^(2*0) + 3 = 2e^0 + 3 = 2 + 3 = 5.
1c) Use the iteration formula:
x1 = x0 - (f(x0) - 15) / f'(x0)
Substitute the values:
x1 = 0 - ((e^(2*0) + 3*0 - 10) - 15) / 5
= -(-10 - 10) / 5
= 20 / 5
= 4.
So, x1 = 4.
1d) We can repeat step 1c) with x1 as the new guess if greater accuracy is desired.
For the calculator method, you can enter the function y = e^(2x) + 3x - 10 into a graphing calculator and find the point of intersection with the line y = 15. Read the x-coordinate of the intersection point to find the x value we are looking for.
Note: These calculations are just examples for illustration purposes. In practice, you may need to perform more iterations or use different initial guesses to achieve the desired level of accuracy.