Question

Determine if the Mean Value Theorem applies to the function on the given interval. If the either applies, find all values of c that are guaranteed by the theorem
f(x)=x^3-x^2-2x on [-1,1]
I do not understand, can some one show me the process?

Answer 1

wat does MVT say ?

Answer 2

First you need to show that the given function satisfies the hypothesis of MVT

Answer 3

It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
\[f'(c)=\frac{ f(b)-f(a) }{ b-a }\]

Answer 4

Yes, since the given function is a simple polynomial, it is continuous in the given interval [-1, 1] and is also differentiable in (-1, 1). So it fits the hypothesis of MVT. we can apply MVT on this function

Answer 5

thats the first part : we have shown that f(x) is continuous + differentiable in given interval

Answer 6

question: how do i do the process to find out if it's differentiable or continuous?

Answer 7

we dont have to prove it rigorously here, we just need show that it satisfies the conditions for MVT loosely

Answer 8

oh ok

Answer 9

next, find f(b), f(a) and f'(c)

Answer 10

and plug them below and solve \(c\)

\(\large f'(c)=\frac{ f(b)-f(a) }{ b-a }\)

Answer 11

\[f'(c)=\frac{ f(1)-f(-1) }{ 1+1} \]

Answer 12

yes, evaluate f(1), f(-1) and f'(c) and plug them above

Answer 13

will the answer be 1

Answer 14

\(\large f'(c)=\frac{ f(1)-f(-1) }{ 1+1}\)

\(\large 3c^2 - 2c-2 =\frac{ -2 -0}{ 1+1}\)

\(\large 3c^2 - 2c-2 =-1\)

\(\large 3c^2 - 2c-1 = 0 \)

\(\large (3c+1)(c-1) = 0\)

\(\large c = -1/3 ~~~ c = 1\)

discard 1 cuz \(1 \not \in (-1, 1) \)

so oly one value for c : -1/3