IVT, MVT & EVT

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

If f(x) is continuous on the interval −2 ≤ x ≤ 5, then there exist a value c such that f(c) ≥ f(x)for all values of x within the interval.

• IVT

• MVT

• EVT

If f(x) is continuous on the interval −2 ≤ x ≤ 5, and f(x) changes from positive to negative within the interval, then there must exist at least one value c such that f(c) = 0.

• IVT

• MVT

• EVT

If f(x) is differentiable on the interval −2 ≤ x ≤ 5, and f '(x) changes from positive to negative within the interval, then there must exist at least one critical point within the interval.

• IVT

• MVT

• EVT

If f(x) is differentiable on the interval −2 ≤ x ≤ 5, and f(-2) = f(5), then there must exist at least one critical point within the interval.

• IVT

• MVT

• EVT

I drove my car continuously down a straight road for 200 miles over a time period of 4 hours. There was at least one point in time when I was going exactly 50 mph.

• IVT

• MVT

• EVT

Tags

CCSS.HSF.IF.B.6

f(x) is continuous and differentiable on the interval −2 ≤ x ≤ 5. There must be at least one point c within the interval where the line tangent to f(x) at x = c is parallel to the line drawn between x = -2 and x = 5.

• IVT

• MVT

• EVT

If f(x) is continuous on a closed interval, then there must exist a maximum and minimum value for the function within that interval.

• IVT

• MVT

• EVT

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

• MVT

• EVT