Question

f(x)=2x^2-7 why the Rolle's theorem don't apply

Answer 1

first, define rolles thrm ....

Answer 2

then check the problem and make sure you havent posted a mistake ...

if that fails, take a screenshot and attach a picture of it to make sure theres no wonky stuff going on in translation ....

Answer 3

I equal f(x) to zero and I have square root of 7/2. f prime gives me zero

Answer 4

not quite sure how that relates ... rolles thrm applies to your stated problem unless youve mistyped something ... and that response is not readable to me

Answer 5

I forgot something it has an interval between [0,4]

Answer 6

so, we want to find 2 points a,b in the interval such that the slope from a to b is 0

\[\frac{f(a)-f(b)}{a-b}=0\]
\[\frac{2a^2-7-(2b^2-7)}{a-b}=0\]
\[\frac{2a^2-7-2b^2+7}{a-b}=0\]
\[\frac{2a^2-2b^2}{a-b}=0\]
\[\frac{2(a^2-b^2)}{a-b}=0\]
so, this is only possible when a^2 - b^2 = 0 giving us a different of squares to factor into

(a+b) or (a-b) = 0; so a=-b, or a=b fit the bill since the denominator is a-b, then a=b is out

Answer 7

is there 2 number in the interval 0 to 4 such that a+b=0?

Answer 8

thank you very much for your help but i don't understand this!

Answer 9

you are looking for 2 number in the interval that have a slope of zero between them .... there are none in that the stated interval therefore Rolles thrm fails.

Rolles thrm: if 2 points in an interval create a slope of 0 ... f(a)=f(b), then there must be a f'(c) = 0 for some c inbetween a and b

Answer 10

since the interval is from 0 to 4, there are no additive inverses that can be used.

Answer 11

so from the original equation you must have an x= to two numbers?

Answer 12

2 different numbers yes

Answer 13

thanks for being so patience with me!

Answer 14

yep .. you never use Rolles thrm but its just there to prove the underling math