Show that the equation #x^6+x^2-1=0# has exactly one positive root. Justify your response. Name the theorems on which your response depends and the properties of #f(x)# that you must use?
1 Answer
Here are a couple of methods...
Explanation:
Here are a couple of methods:
Descartes' Rule of Signs
Given:
#f(x) = x^6+x^2-1#
The coefficients of this sextic polynomial have signs in the pattern
We also find:
#f(-x) = f(x) = x^6+x^2-1#
which has the same pattern of signs
Turning points
Given:
#f(x) = x^6+x^2-1#
Note that:
#f'(x) = 6x^5+2x = 2x(3x^4+1)#
which has exactly one real zero, of multiplicity
Since the leading term of
We find