Exam Hub Rendering Issues (copy)

This statement is correct. For Rolle's Theorem to apply, the function must indeed be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). These conditions ensure that the function does not have any breaks, jumps, or sharp corners in the interval, which are necessary conditions for the existence of a point c where the derivative f'\(c) = 0.

renders the first three latex in the editor but not in the rendering

Which of the following is the correct antiderivative of \frac{1}{\sqrt{x}}?

Square roots aren’t rendering properly in the editor

Rank the following functions in order of increasing difficulty to differentiate using the chain rule: f(x) = (3x^2 + 2)^5, g(x) = \sin(5x^3), h(x) = \e^{2x^2 + 3x}, j(x) = \sqrt{4x + 1}.

The one \e messes up everything when actually it could render a lot better

Given the function y = \sqrt{x^3 + 3x^2 + x}, rank the following steps in the correct order for applying the chain rule to find dy\dx: A) Differentiate the outer function, leaving the inside function alone, B) Multiply by the derivative of the inside function, C) Identify the inside and outside functions, D) Simplify the derivative expression.

Again renders the first one and semi renders the second but doesn’t work for either term on rendering

Rank the following functions by the likelihood of having a jump discontinuity at x = 2: f(x) = \frac{x^2 - 4}{x - 2}, g(x) = \piecewise{{x + 2, x < 2}, {x^2, x >= 2}}, h(x) = x^3, j(x) = \frac{1}{x - 2}.

I. The definite integral \int_{0}^{\pi} \cos(x) dx equals 0.II. The definite integral \int_{1}^{2} (3x^2 + 2x + 1) dx equals \frac{17}{3}.III. The definite integral \int_{-2}^{2} x^3 dx equals 0.Which of the following statements are true?

Doesn’t have new lines