Answer to Question #327244 in Calculus for Ayeeshaaaa

Question

1. Prove that h(x) = x² - x - 6  has a root on [ -4 , 0 ]

Answer 1

ANSWER

First note, that the function is defined and continuous on $[-4,0]$ . $h(-4)=(-4)^{2}-(-4)-6=16+4-6=14>0$ , $h(0)=-6<0$ . So $h(0)<0<h(-4).$ By the Intermediate Value Theorem there exists a number $c\in(-4,0)$ such that $h(c)=0$ .

Note: the root of the equation is easy to calculate by solving the equation $x^2-x-6=0.$

$c _{1}=\frac{1+\sqrt{1+24}}{2}= \frac{1+5}{2}=3 \notin( -4,0)$, $c _{2}=\frac{1-\sqrt{1+24}}{2}= \frac{1-5}{2}=-2 \in (-4,0)$ .

$c=c_{2}$.