IVT, EVT, MVT, and Rolle's Theorem

Rolle's Theorem

Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number c in (a, b) such that f'(c)=0.

Let g(x) = 2ˣ Can we use the mean value theorem to say the equation g'(x)=16 has a solution where 3 < x < 5?

No, since the average rate of change of g over the interval 3 ≤ x ≤ 5 isn't equal to 16.

Let h be a continuous function on the closed interval [1, 5], where h(1) = 6 and H(5) = -3 Which of the following is guaranteed by the Intermediate Value Theorem? (a) h(c) = -2 for at least one c between 1 and 5 (b) h(c) = 4 for at least one c between -3 and 6 (c) h(c) = -2 for at least one c between -3 and 6 (d) h(c) = 4 for at least one c between 1 and 5

(a) h(c) = -2 for at least one c between 1 and 5

Intermediate Value Theorem (IVT)

If f is continuous on [a,b] and N is any number between f(a) and f(b), then there exists at least one number c in the open interval (a,b) such that f(c)=N.

Mean Value Theorem (MVT)

If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one number c in the open interval (a,b) such that f'(c)= (f(b)-f(a))/(b-a).

Extreme Value Theorem (EVT)

If f is continuous on a closed interval [a,b], then f has both a minimum and a maximum on the interval.

Let g(x) = tan(x) Is it possible to use the intermediate value theorem to say the equation g(x) = 0 has solution where (π/4) ≤ x ≤ (3π/4)?

No, since the function is not continuous on that interval.

Let f(x) = x³+6x²+6x and let c be the number that satisfies the Mean Value Theorem for f on the interval [-6, 0] What is c?

c = -4

Let f(x) = √(4x - 3) and let c be the number that satisfies the Mean Value Theorem for f on the interval 1 ≤ x ≤ 3 What is c?

c = 1.75

Let g(x) = √(x³ + 8x), for the interval (0,100) Where does g have critical points?

g has no critical points.

Let g(x) = 5 - eˣ Give a formal justification for the fact that the equation g(x) = 0 has a solution where 1 ≤ x ≤ 4.

g is defined for all real numbers, and exponential functions are CONTINUOUS at all points in their domains. 0 is BETWEEN g(1) and g(4). Therefore, according to the IVT, g(x) = 0 must have a solution somewhere between x = 1 and x = 4.

Let h(x) = (eˣ)/(x-1), for the interval (0,4) Where does h have critical points?

x = 2