Question

the equation x^3 -x-5=0 has one solution for x between -2 and 2. what is the solution?

how do I get this

Answer 1

The only real root is x = 1.0416
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php

Answer 2

1.90416

Answer 3

how do you get that without using a calc?

Answer 4

I am also having trouble with this problem

for a differentiable function it is known that f(3)=5 and f'(3)=-2 use differentials to get the approx. value of f(3.02)

Answer 5

I can do better than that. its between 1 and 2. Ever hear of the intermediate value theorem?
looking at 1.8=x
then it is still negative (y=x^3-x+5), so it is now between 1.8 and 2
looking at 1.9, same logic, it is now bounded by 1.9<x<2.0
now looking at 1.95, it is positive, so the 1.9<x<1.95, positive, 1.9<x<1.93
looking at x=1.92, too high
looking at x=1.91, too high
looking at x=1.905, a hair to high
looking at x=1.9042, that is really close. How accurate do you want it.

Check out the intermediate value theorem, and Newton's method (more complicated by hand).

Answer 6

Did you click on the polynomial calc link?

You can use Newton's method.
I will give you a link in a minute.
Oh, no need, this next problem is in fact Newton's method.

Draw a sketch
f(3) = 5
Put an x at (3,5)
slope ay (3,5) = -2
f(x+dx) = f(x) + slope * dx
f(3 + .02) = f(3) -2 (.02)
f(3.02) = 5 -.04
= 4.96
Now back to the earlier problem which is this same trick in a minute

Answer 7

Newton's method for finding zeros
f(x) = x^3 -x-5
pick an x in the range, say x = 2

f(2) = 1
f'(2) = 3 x^2 - 1 = 11
mark that point (2,1) and slope of 11 through it

now that slope will hit the x axis at
x + y / dy/dx
or
2 + (-1)/11 = 1.909
now repeat again and again until you
converge
f(1.909) =
f'(1.909 ) = etc

Answer 8

sorry, x - y / dy/dx

Answer 9

http://archives.math.utk.edu/visual.calculus/3/newton.5/