Rolle's Theorem tells us that when a function is continuous on a closed interval and differentiable on the open interval, and the endpoint values are equal, there must be an interior point where the tangent is horizontal. In other words, there exists a c in the open interval such that the derivative at c is zero. Why that must be the case: a continuous function on a closed interval attains its maximum and minimum there. If the endpoints have the same value, at least one interior point must achieve a local maximum or minimum. At any interior point where the function has a local extremum, Fermat’s theorem says the derivative must be zero; the tangent line is horizontal, so f'(c) = 0. The other possibilities—derivative positive, derivative negative, or derivative undefined—contradict either the interior extremum condition or the differentiability requirement in the interval. Since the theorem guarantees differentiability on the open interval, the derivative cannot be undefined at c, and the local extremum forces the derivative to be zero rather than strictly positive or negative. Note: the statement relies on the endpoint condition f(a) = f(b); without that, the conclusion need not hold.

When x gets really large, compare how fast each function goes to infinity. Exponential growth dominates every polynomial: for any base a > 1 and any positive n, the ratio a^x / x^n tends to infinity as x → ∞. So exponentials outrun polynomials. Next, polynomials outrun logarithms: for any n > 0, x^n / log x → ∞ as x → ∞. Therefore, the fastest to slowest among these is exponential, then polynomial, then logarithmic growth. The standard ranking from fastest to slowest is exponential, polynomial, logarithmic.

When a region is revolved around an axis and there is a hollow inside, each cross-section perpendicular to the axis is a washer: an outer disk of radius R(x) with a hole of radius r(x). The area of that washer is π[R(x)]^2 − π[r(x)]^2. To get the volume, you sum these cross-sectional areas from a to b, which gives V = ∫_a^b [π(R(x)^2) − π(r(x)^2)] dx. This is the same as V = π ∫_a^b [R(x)^2 − r(x)^2] dx, i.e., the outer disk’s volume minus the inner hole’s volume. Expressed differently, you can see it as the difference of two separate disks: π ∫_a^b R(x)^2 dx minus π ∫_a^b r(x)^2 dx. This matches the correct idea: the outer radius contributes the full disk area, the inner radius subtracts the hollow part. The other forms miss a piece: dropping the inner subtraction inside the integral omits the hole, and dropping the π factor would misscale the volume. If the inner radius were zero, it would reduce to the disk method.

This uses the second derivative test for a critical point. If f'(a) = 0, a is a candidate for a local extremum. When f''(a) > 0, the graph is concave up at a, like a bowl, so values of f nearby are greater than f(a) and the tangent is horizontal there. That makes a a local minimum. If f''(a) < 0, it would be a local maximum, and if f''(a) = 0 the test is inconclusive. An inflection point would require a change in concavity, which does not occur here since the second derivative is positive. So the point a is a local minimum.