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SOLUTION 11: We are given the equation
x4x+1=x2+5 ⟶ x4=(x+1)(x2+5)
⟶ x4=x3+x2+5x+1
⟶ x4−x3−x2−5x−1=0
Let function
f(x)=x4−x3−x2−5x−1 and choose m=0
This function is continuous for all values of x since it is a polynomial. We now need to search for an appropriate interval satisfying the assumptions of the Intermediate Value Theorem. By trial and error, we have that
f(0)=(0)4−(0)3−(0)2−5(0)−1=−1<0 and f(3)=(3)4−(3)3−(3)2−5(3)−1=29>0
so that f(0)=−1<m<29=f(3)
i.e., m=0 is between f(0) and f(3).
The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number c in the interval [0,3] which satisfies
f(c)=m
i.e.,
c4−c3−c2−5c−1=0
and the equation is solvable.
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