approximate each real zero as a decimal to the nearest tenth

Question

Given f(x)=6x^4-7x^3-23x^2+14x+3, approximate each real zero as a decimal to the nearest tenth.

Answer 1

approximate the zero of the function y=x^2+6x_2 to the nearest tenth

Answer 2

Daniel,

 

This is one of those functions you almost have to graph in order to see what's going on.  Doing so will show that there are real zeros in the intervals [-2<x<-1.5], [-0.05<x<0], [0.5<x<1] and [2<x<2.5].  As to how to solve for these zeros, we could use the method of Newton

 

x₁ = x₀ - f(x₀)/f'(x₀) 

 

where x₀ is your "first guess" as to the real zero's value, f(x₀) is f(x) evaluated at that point, and f'(x₀) is the derivative of f(x) evaluated at x₀.

 

For example, we know from our graph that there is a real zero between -2 and - 1.5.  Let's push our luck and assume that that value is -1.75.

 

Then

 

x₁ = (-1.75) - {[6(x₀)⁴ - 7(x₀)³ - 23(x₀)² + 14(x₀) + 3]/[24(x₀)³ -21(x₀)² -46(x₀) + 14]}

 

x₁ = (-1.75) - (1.85156)/(-98.4375) = -1.73119

 

Now you can take the value for x1 and use it to find the next approximation, x2, by going through the process all over again.

 

x₂ = x₁ - [f(x₁)/f'(x₁)]

 

To save you some trouble, I evaluated the original expression using a TI-89 Titanium calculator and here are the values I found:

 

x = -1.73073, or -0.169749, or .727385 or 2.33976