f(x)=x^3+5x-4
f(0)=-4
f(1)=2
Since f(x) is continuous (as are all polynomials), there exists a value c such that f(c)=0 (-4≤0≤2, and 0≤c≤1) by the Intermediate Value Theorem.
X^3+5x-4=0 Show this equation has a solution between x=0 and x=1.
(The question is two marks)
3 answers
I found out how to answer the question myself but for others wondering how, I solved it by subbing in the numbers and stating there is a change of sign. This would ensure the full two marks.
F(x)=x^3+5x-4
F(0)=(0)^3+5(0)-4
F(0)= -4
F(1)=(1)^3+5(1)-4
F(1)=2
There's a change of sign therefore the solution will lie between 0 and 1.
F(x)=x^3+5x-4
F(0)=(0)^3+5(0)-4
F(0)= -4
F(1)=(1)^3+5(1)-4
F(1)=2
There's a change of sign therefore the solution will lie between 0 and 1.
Exactly, we're looking for the change of sign which implies the presence of a root. It may be intuitive, but this fact is dependent on the Intermediate Value theorem, as stated above.