The only real root is x = 1.0416
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
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
9 answers
1.90416
how do you get that without using a calc?
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)
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)
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).
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).
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
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
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
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
sorry, x - y / dy/dx
http://archives.math.utk.edu/visual.calculus/3/newton.5/
1 answer