0.45 1. The rate at which the height of the water increases when the water is 10ft high is 7 feet per minute. 2. The linearization of the function f (x) = vex -1 at * = In(5) is L(x) = 8x - 5in(5 + 2 3. For the function f () = a" - 1" + 2" on the interval [0, 2], the absolute minimum is 0 at =, and the absolute maximum is 4 at " = 2. (PLEASE SEE DETAILED INFORMATION ON THE EXPLANATION PROVIDED.) Explanation: 1. The volume of a cone in terms of its height ^ is given by: V = mr2h Since the radius " of the water level and the height " of the water level maintain the same ratio as the cone itself (due to similar triangles), we have? = 3. Substituting this into the volume formula, we get: V = ;T(3h)2h = 27Th3 We take the derivative with respect to time * to relate the change in volume at to the change in height dt : dh dv = Tah dt Given that water is being pumped at a rate of 2020 cubic feet per minute at = 2, we solve for at when h = 10feet dV 20 = 71 2 (10) 2 dh dt dh 20 d+ T . (10) 2 dh 0.45 d+2. For linearization, we find the first derivative of flz) = Ve - 1 evaluate it at = ln(5), and use the point-slope form of the line: ff(z) _ 2\\/(5127_16:: z = In(5) Evaluating at (in(5)) = 5 e WO Since f'(In(5)) = v5 1= 2, the linearization L(%) at & = In(5) js. L(z) = f(In(5)) + f'(In(5))(z In(5)) L(z) =2+ 2(z In(5)) L(m):%msmf,'(s)+2 . . .. . . f($)=$3_1$2+$ . . 3. To find the absolute maxima and minima of a continuous function 1 2** on a closed interval [0,2][0,2], we find the derivative and solve for critical points where the derivative is zero or undefined: f'(z) = 3% fz + ! Setting f ('7') =0 and solving within the interval [0, 2], we get the critical points. Then, we evaluate the function at these critical points as well as at the endpoints of the interval to find the absolute extrema. Find that f(O) =0 and f(Z) = 4, and since the function is a cubic polynomial (which has no discontinuities or sharp corners), these are the absolute minimum and maximum values on the interval [0, 2], respectively