How do you verify the intermediate value theorem over the interval [0,3], and find the c that is guaranteed by the theorem such that f(c)=4 where #f(x)=x^3-x^2+x-2#?

2 Answers
Aug 14, 2017

#f# is a polynomial so #f# is continuous on the interval #[0,3]#.

#f(0) = -2# and #f(3) = 19#

#4# is between #f(0)# and #f(3)# so IVT tells us that there is a #c# in #(0,3)# with #f(c) = 4#

Finding the #c# requires solving

#x^3-x^2+x-2 = 4#. Which is equivalent to

#x^3-x^2+x-6 = 0#

Possible rational zeros are #+-1#, #+-2#, #+-3#, and #+-6#.

Testing shows that #2# is a solution.

So #c = 2#

Aug 14, 2017

The value of #c in (0, 3)# and is #c=2#

Explanation:

The intermediate value theorem states that if #f(x)# is a continuous function on the interval #[a,b]#, then there is a number #p# between #f(a)# and #f(b)#, #(f(a)!=f(b))# such that there is a number # c in (a,b)# such that #p=f(c)#

Here, #f(x)=x^3-x^2+x-2#, which is a polynomial function continuous on the Interval #[0,3]#

#f(0)=0-0+0-2=-2#

#f(3)=3^3-3^2+3-2=27-9+3-2=19#

#f(c) in (f(0, f(3)))#

#f(c)=4# such that #c in (0,3)#

Therefore,

#f(c)=c^3-c^2+c-2=4#

#=>#, #c^3-c^2+c-6=0#

#f(2)=2^3-2^2+2-2=8-4+2-2=4#