# calculus

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.

1. 👍
2. 👎
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1. 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

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2. 👎
👤
Reiny
2. 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

1. 👍
2. 👎
👤
Damon

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