Max Number of Real Zeros
- Thread starter harpazo
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Looking at:
[MATH]f(x)=5x^4+2x^2-6x-5[/MATH]
We see there is one sign change in the coefficients, which means there is a maximum of one positive zero. Next:
[MATH]f(-x)=5x^4+2x^2+6x-5[/MATH]
We see there is one sign change in the coefficients, which means there is a maximum of one negative zero. And so we conclude there is a maximum of two zeroes, one negative and one positive.
[MATH]f(x)=5x^4+2x^2-6x-5[/MATH]
We see there is one sign change in the coefficients, which means there is a maximum of one positive zero. Next:
[MATH]f(-x)=5x^4+2x^2+6x-5[/MATH]
We see there is one sign change in the coefficients, which means there is a maximum of one negative zero. And so we conclude there is a maximum of two zeroes, one negative and one positive.
harpazo
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Mark,
Can we also say that the simplest maximum number of real zeros is equal to the highest power of the given function?
For example:
Say our function is given in the form
f(x) = mx + b. Can I say that f(x)
has a maximum of one (x is x^1) real zero?
What about f(x) = ax^2 + bx + c?
Can I safely say that f(x) has a maximum of two (x^2) two real zeros?
Only for certain types of functions. If your f(x) is a polynomial, it will be true that the maximum number of real zeros is the same as the highest power. The Fundamental Theorem of Algebra states that an \(n^{th}\) degree polynomial has exactly \(n\) roots. If all of these roots are real, then the maximum number of real roots of f(x) is \(n\). However, this is not always true for other types of functions. Consider, for example, \(g(x) = x^2 - \sin(5x)\). This function has four real roots, which violates your hypothesis that the highest power dictates the maximum number of real roots.
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