Question

Use the Intermediate Value Theorem to prove that the function f (x) = x^2-4x+3 intersects the x-axis on the interval [0,2].
Can we use the same theorem to say the same about the function
g(x) =x^2-2x+1/x-1?
ANSWER EITHER YES OR NO ?
Not sure how to go about doing this ? do you fill in 0,1... and so on for f(x) until you get a positive and Negative ?

Answer 1

if the graph passes x-axis it will have a zero in the interval and on both sides of zero the sign of function will change

check f(0) and f(2)
if the sign changed means there is a zero in there

Answer 2

yes i did that previous and the signs changed which means a negative by a positive is a negative therefore its continuous . but came we same the same for the second part or how do you go about solving it ??? answer only allows me to say yes or no ??

Answer 3

you can not use the intermediate value therom on the second part becuase it is not continuous on the interval from [0.2]

Answer 4

ok cheers

Answer 5

f(x) is continuous for all x, hence it is continuous on interval [0,2]. So we can apply the IVT
-1 = f(2) < 0 < f(0) = 3
therefore there must be a c in [0,2] such that f(c)=0

g(x) =(x^2-2x+1)/(x-1) is not continuous when x=1
so the IVT doesn't apply over any interval containing x=1