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

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

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

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Austin, TX

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

Aliquippa, PA

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

x_{1} = x_{0} - f(x_{0})/f'(x_{0})

where x_{0} is your "first guess" as to the real zero's value, f(x_{0}) is f(x) evaluated at that point, and f'(x_{0}) is the derivative of f(x) evaluated at x_{0}.

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

x_{1} = (-1.75) - {[6(x_{0})^{4} - 7(x_{0})^{3} - 23(x_{0})^{2} + 14(x_{0}) + 3]/[24(x_{0})^{3} -21(x_{0})^{2} -46(x_{0}) + 14]}

x_{1} = (-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.

x_{2} = x_{1} - [f(x_{1})/f'(x_{1})]

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

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