MVT, IVT, EVT

Which value is guaranteed by the IVT for the function given: f(x) = x² - 6x + 8 on [0,3]

• f(c) = 10

• f(c) = 4

• f ' (c) = 3

• f ' (c) = 7/3

Which value is guaranteed by the MVT for the function given: f(x) = x² - 6x + 8 on [0,3]

• f(c) = 10

• f(c) = 4

• f ' (c) = -3

• f ' (c) = 7/3

Does the IVT apply to the function on the given interval?

f(x) = (x - 2)/(x² - 16) on [2 , 6]

• Yes

• No

Find the value of c guaranteed by the IVT for the function on the given interval:

f(x) = x² + x - 1 on [0,5] : f(c) = 11

• c = -4

• c = 1

• c = 3

• c = 12

Can the MVT be applied to the function on the interval?

f(x) = (-x² + 9)/(4x) on [1,3]

• Yes

• No

Find all values of c that satisfy the MVT for the function on the given interval

f(x) = (-x² + 9)/(4x) on [1,3]

• (-3)^(1/2)

• -1

• 1

• 3^(1/2)

Find all values of c that satisfy the MVT for the function on the given interval

f(x) = -x² + 8x - 17 on [2,6]

• c = 2

• c = 3

• c = 4

• c = 6

Tags

CCSS.HSA.REI.D.11

Does the EVT apply to the function on the given interval?

f(x) = x²/(3x - 6) on [3,6]

• Yes

• No

f(x) is continuous and differentiable on the interval −2 ≤ x ≤ 5. If f(-2) = 5 and f(5) = -2, then there must exist at least one value c in the interval such that f '(c) = -1.

• IVT

• EVT

• MVT

If f(x) is continuous on the interval −2 ≤ x ≤ 5 and f(-2) = 5 and f(5) = -2, then there must exist at least one value c in the interval such that f(c) = 3.

• IVT

• MVT

• EVT