This site is best viewed with Mozilla Firefox.

Please install Math Player to see the Math Symbols properly

Attempt following question by selecting a choice to answer. c Find the value in the interval, which satisfies the Mean Value Theorem for the function

f (x ) = x - x ^{3} on

[- 2, 1]. f c A.

c 1

B.

c 0

C.

c 6

D.

c - 1

E.

c 1 3 c
D

Step 1: f (x ) = x - x ^{3} [Write the function.] Step 2: Since f (x ) is continuous in [- 2, 1] and differentiable in (- 2, 1), Mean Value Theorem is applicable. Step 3: f (- 2) = - 2 - (- 2)^{3} = - 2 + 8 = 6[Find f (- 2).] Step 4: f (1) = 1 - (1)^{3} = 1 - 1 = 0[Find f (1).] Step 5: f ′(x ) = 1 - 3x ^{2} [Find f ′(x ).] Step 6: f ′(c ) = 1 - 3c ^{2} [Find f ′(c ).] Step 7: By Mean Value Theorem, there exists c ∈ (- 2, 1) such that f ′(c ) = f ( 1 ) - f ( - 2 ) 1 - ( - 2 ) . Step 8: 1 - 3c ^{2} = 0 - 6 3 Step 9: 1 - 3c ^{2} = - 2 Step 10: c = - 1, 1[Solve for c .] Step 11: 1 Ï (- 2, 1), but c = - 1 ∈ (- 2, 1)
Step 12: At x = - 1 in (- 2, 1), the function f (x ) satisfies the Mean Value Theorem.