Answer:
c = 1
Explanation:
Given:-
- The given function f(x) is:
f(x) = 3x^2 - 2x + 1 , [ 0 , 2 ]
Find:-
Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem.
Solution:
- The mean value theorem states that if a function f(x) is differentiable over the range [ x1 , x2 ] , then there exist a value c within the range [ x1 , x2 ]. Such that:
f'(c) = [ f(x2) - f(x1) ] / [ x2 - x1 ]
- Note: The right hand side of above theorem expresses the " Secant " line.
- We see that the function f(x) is a polynomial function with degree 2 which is continuous and differentiable over the entire interval of real numbers R. For which the differential of the given function is:
f'(x) = 6x - 2
- It exist for all real value of x and is continuous ( Linear Line ).
- It satisfies the hypothesis of the mean value theorem. So our function f(x) to be differentiable over the range [ 0 , 2 ]. then there exist a value c within the range [ 0 , 2 ]
f'(x) = [ f(2) - f(0) ] / [ 2 - 0 ]
f'(x) = [ 3(2)^2 - 2(2) + 1 - 1 ] / [ 2 - 0 ]
f'(x) = [ 8 ] / [ 2 ] = 4
f'(c) = 6c - 2 = 4
c = 6 / 6 = 1
- Hence, the required value of c = 1.