Answer

To find the extreme values of a function f(x.y) on a curve x-x(t), y y(t), treat f as a function of the single variable t and use the chain rule to find where df/dt is zero. in any other single-variable case, the extreme values of f are then found among the values at the critical points (points where df/dt is zero or fails to exist), and endpoints of the parameter domain. find the absolute maximum and minimum values of the following function on the given curves. use the parametric equations x=2cos t, y 2 sin t functions: curves: i) the semicircle x4,y20 i) the quarter circle x2+y-4, x20, y20 b, g(x,y)=xy

Complete the solution of the equation. find the value of y when x equals -5. -4x - 8y = -52

Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x^2 + 3x + 4, [−1, 1] no, f is continuous on [−1, 1] but not differentiable on (−1, 1). no, f is not continuous on [−1, 1]. yes, f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous and differentiable on . there is not enough information to verify if this function satisfies the mean value theorem. yes, it does not matter if f is continuous or differentiable; every function satisfies the mean value theorem.