Question

if f(x)= x^3-x^2+x show that there is a number c such that f(c)=10

Answer 1

really? oh i bet this is a "intermediate value" theorem problem isn't it?

Answer 2

the one after this is, and im reading on it as we speak.. im just a bit lost

Answer 3

i guess this is one as well, but it doesnt specify.

Answer 4

is*

Answer 5

really not much to show because
\[f(x)=x^3-x^2+x\] is continuous and has range all real numbers. so not only does it equal 10 somewhere but it equal every number somewhere

Answer 6

if you want to be a little more specific (which is silly) you could say that
\[f(0)=0, f(3)=21\] and since
\[0<10<21\] there must be a number c in the interval (0,3) for which
\[f(x)=10\]

Answer 7

"intermediate value theorem" basically says that a continuous function cannot skip values. so if
\[f(x_1)=a\] and
\[f(x_2)=b\] then for any number with c
\[a<c<b\] there is an
\[x_3\] with
\[f(x_3)=c

Answer 8

of course i am assuming that
\[a<b\] othewise flip the inequality

Answer 9

you just helped me on my next problem lol but thats it ? just plug in a number for example is has (0,1) for the function is x^3-3x+1

Answer 10

not sure what the question is. are you given a specific value?

Answer 11

it says use the intermediate value theroem to show that there is a root of the given equation in the specific interval. x^3-3x+1=0 , (0,1)

Answer 12

right

Answer 13

the whole root thing kind of threw me off...

Answer 14

plug in 0 for x, see that you get some negative number

Answer 15

plug in 1 for x, see that you get some positive number

Answer 16

and since 0 is between whatever negative number you get and whatever positive number you get, you can conclude that there is some c in (0,1) with f(c)=0

Answer 17

"root" in this case means a solution to
\[x^3-3x+1=0\]

Answer 18

clear or no?

Answer 19

yes, its clear, so i did and for 0 i got 1 and for 1 i got -1. so c would be between -1 and 1?