Use the intermediate value theorem to show that there is a root (solution) of the given equation in the interval (1,2)

x^3-2x=3

Question

Use the intermediate value theorem to show that there is a root (solution) of the given equation in the interval (1,2)

x^3-2x=3

anonymous · Answer

put $f(x)=x^3-2x$ and compute $f(1)$ and $f(2)$
one is  since $f(1)<3$ and $f(2)>3$ there must be a number $c$ in the interval (1,2) where $f(c)=3$

anonymous · Answer

k i would get f(1)=3 and f(2)=0

anonymous · Answer

Well 1<2 a<b
f(x)=x^3-2x-3=0
You have to plug 1 and 2 into the equation
(1)^3-2(1)-3
1-2-3
1-5
-4

(2)^3-2(2)-3
8-4-3
8-7
1
Because f(1)<0 and f(2)>0 and f is continuous, therefore there exist a zero in the interval 1 and 2.

anonymous · Answer

minus 3 from both sides of the equation

anonymous · Answer

k didn't c that