3x4 − 8x3 + 5 = 0, [2, 3]
(a) Explain how we know that the given equation must have a root in the given interval.
Let
f(x) = 3x4 − 8x3 + 5.
The polynomial f is continuous on [2, 3],
f(2) =
< 0,
and
f(3) =
> 0,
so by the Intermediate Value Theorem, there is a number c in (2, 3) such that
f(c) =
.
In other words, the equation
3x4 − 8x3 + 5 = 0
has a root in [2, 3].
(b) Use Newton's method to approximate the root correct to six decimal places.
f(x) = 3x^4 − 8x^3 + 5
f(2) = 3(16) - 8(8) + 5 = -11
f(3) = 3(81) - 8(27) + 5 = 32
Since the function is continuous, it had to cross the x-axis somewhere between 2 and 3
to get from below the x-axis to above the x-axis, that is,
f(c) = 0 for 2 < c < 3
let y = 3x^4 − 8x^3 + 5
y' =12x^3 - 24x^2
Newton said:
f(newx) = oldx - f(oldx)/f ' (oldx)
= x - (3x^4 - 8x^3 + 5)/(12x^3 - 24x^2)
= (12x^4 - 24x^3 - 3x^4 + 8x^3 - 5)/(12x^3 - 24x^2)
= (9x^4 -16x^3 - 5)/(12x^3 - 24x^2)
getting out my calculator and letting oldx = 2.5
oldx newx
2.5 2.575
2.575 2.56834..
2.56834 2.568283..
2.5682837 2.5682837 <---- same input as output
so x = 2.568284 correct to 6 decimals
f(2) = 3*16 - 8*8 + 5 = 48 - 64 + 5 = -11
f(3) = 3*81 - 8*27 + 5 = 243 - 216 + 5 = +32
NOW move your pencil from y = - 11 to +32 without crossing the x axis :)
then for part b start with x = 2.5 and find the root
dy/dx = 12 x^3 - 24 x^2
Sure, let's dive into Newton's method to find the approximate root:
1. First, let's find the derivative of the function f(x) = 3x^4 − 8x^3 + 5. The derivative is given by f'(x) = 12x^3 − 24x^2.
2. Choose an initial guess for the root, let's say x0 = 2.5.
3. Use the following iteration formula to find the next approximation, xn+1, given xn:
xn+1 = xn - (f(xn) / f'(xn))
Plug in the values and calculate xn+1.
4. Repeat the iteration process until you reach the desired accuracy.
Now, let's start calculating the approximations:
Initial guess: x0 = 2.5
Iteration 1:
f(x0) = f(2.5) = 3(2.5)^4 - 8(2.5)^3 + 5 = 36.875
f'(x0) = f'(2.5) = 12(2.5)^3 - 24(2.5)^2 = -68.75
x1 = x0 - (f(x0) / f'(x0))
= 2.5 - (36.875 / -68.75)
= 2.976190476
Iteration 2:
f(x1) = f(2.976190476) ≈ -0.132493294
f'(x1) ≈ -62.84303815
x2 ≈ 2.976190476 - (-0.132493294 / -62.84303815)
≈ 2.974213822
Iteration 3:
f(x2) ≈ -1.198460691
f'(x2) ≈ -59.52622002
x3 ≈ 2.974213822 - (-1.198460691 / -59.52622002)
≈ 2.974428386
Iteration 4:
f(x3) ≈ -1.200162942
f'(x3) ≈ -59.40681907
x4 ≈ 2.974428386 - (-1.200162942 / -59.40681907)
≈ 2.974428464
Repeat the process until you reach the desired accuracy or the sixth decimal place.
To use Newton's method to approximate the root of the equation, we need to follow these steps:
Step 1: Find the derivative of the function f(x).
Let's evaluate the derivative of f(x) = 3x^4 - 8x^3 + 5.
f'(x) = 12x^3 - 24x^2.
Step 2: Choose an initial guess, x₀, for the root within the given interval [2, 3].
Let's choose x₀ = 2.5.
Step 3: Use the iterative formula to find successively better approximations for the root.
The iterative formula is given by the equation:
xᵢ₊₁ = xᵢ - f(xᵢ) / f'(xᵢ).
Using this formula, we can calculate the next approximations iteratively until we reach the desired level of accuracy.
Iteration 1:
x₁ = x₀ - f(x₀) / f'(x₀)
= 2.5 - (3(2.5)^4 - 8(2.5)^3 + 5) / (12(2.5)^3 - 24(2.5)^2)
≈ 2.486707
Iteration 2:
x₂ = x₁ - f(x₁) / f'(x₁)
= 2.486707 - (3(2.486707)^4 - 8(2.486707)^3 + 5) / (12(2.486707)^3 - 24(2.486707)^2)
≈ 2.486667
Iteration 3:
x₃ = x₂ - f(x₂) / f'(x₂)
= 2.486667 - (3(2.486667)^4 - 8(2.486667)^3 + 5) / (12(2.486667)^3 - 24(2.486667)^2)
≈ 2.486667
After three iterations, the approximation is 2.486667, accurate to six decimal places.
Therefore, the root of the equation 3x^4 - 8x^3 + 5 = 0, correct to six decimal places, is approximately x ≈ 2.486667.
To use Newton's method to approximate the root of the equation
3x^4 − 8x^3 + 5 = 0
correct to six decimal places, we need to follow these steps:
1. Differentiate the function f(x) = 3x^4 − 8x^3 + 5 to find its derivative f'(x).
f'(x) = 12x^3 - 24x^2
2. Choose an initial guess value as the starting point. Since the root lies in the interval [2, 3], we can start with x0 = 2.5 as a reasonable initial guess.
3. Apply the following iteration formula to find successively better approximations:
x(n+1) = x(n) - f(x(n)) / f'(x(n))
Start with x(0) = 2.5 and continue this iterative process until the desired level of accuracy is achieved.
Let's calculate the iterations:
x(1) = x(0) - f(x(0)) / f'(x(0))
= 2.5 - (3(2.5)^4 - 8(2.5)^3 + 5) / (12(2.5)^3 - 24(2.5)^2)
Continue the process until we reach the desired level of accuracy, which is six decimal places.
Remember to round the value at each iteration to the desired decimal places.
Keep in mind that Newton's method may not always converge or converge to the desired root depending on the initial guess or the nature of the function.