Given: Function: $f(x)=x^2-2x-3$ $[1,4]$
Question: Show that the function has exactly one root in $(1,4)$
My Answer: The function $f(x) = x^2-2x-3$ has one root in [1,4]
$f(1)=1^2-2(1)-3=1-2-3=-4$
$f(4)=4^2-2(4)-3=16-8-3=5$
Since $f$ is continuous, $f(1)<0$, and $f(4)>0$, by the Intermediate Value Theorem, there must be a value between 1 and 4, say $c$, such that $f(c)=0$. So we have at least one root.
To prove that there is the function has exactly one root in $(1,4)$, Consider the following. Remember that Rolle’s theorem states that if a function is continuous on [a,b] and differentiable on (a,b) where $f(a)=f(b)$ then there exists c in $(a,b)$ such that $f’(c)=0$
Assume that this function has 2 roots : $f(a)=f(b)=0$. Then there exists c in (a,b) such that $f’(c)=0$. However, $f’(x)=2x-2>0$ is always positive so there exists no c such that f’(c)=0. Therefore, there is exactly one root since there cannot be two or more roots. Hence $f(x) = x^2-2x-3=0$ only has one exactly root.
Check: Could anyone verify this work and see if my answer is correct or wrong? If it is wrong, can you help me to correct it?
Thank You.
