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
The only real root is x = 1.0416
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
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)
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).
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
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
sorry, x - y / dy/dx
http://archives.math.utk.edu/visual.calculus/3/newton.5/
To find the solution to the equation x^3 - x - 5 = 0, you can use a numerical method called the Bisection Method.
Here's how you can do it step by step:
1. Begin by narrowing down the interval where the solution lies. Since it is given that the solution is between -2 and 2, you can start with this interval.
2. Calculate the value of the function at the boundaries of the interval. Plug in -2 and 2 into the equation:
For x = -2: (-2)^3 - (-2) - 5 = -11
For x = 2: 2^3 - 2 - 5 = 1
3. Check for a sign change. In this case, there is a sign change between -2 and 2 since the function value is negative at -2 and positive at 2.
4. Choose the midpoint of the interval (-2, 2) and evaluate the function value at that point.
Midpoint = (-2 + 2)/2 = 0
For x = 0: 0^3 - 0 - 5 = -5
5. Again, check for a sign change. In this case, there is a sign change between 0 and -2.
6. Narrow down the interval based on the sign change. Since there is a sign change between -2 and 0, the solution lies in this interval.
7. Repeat steps 4 to 6 iteratively, narrowing down the interval further until you obtain the desired level of precision.
Using this method, you can continue these steps until you find the approximate solution within the given interval.