Consider g(x) = 5x - x^2. The Extreme Value Theorem guarantees that g attains ma

Question

Consider g(x) = 5x - x^2. The Extreme Value Theorem guarantees that g attains maximum and minimum values of [1, 3]. Use calculus to find the exact max, and min values of g on this interval. Evaluate integral^1-1(6x - x^2) dx. Use the Mean Value Theorem of the integral calculus to determine the value of the function g on [1, 3]. F(x) = 1/x. Use the definition of derivative to compute f(5) and determine an equation of the line tangent to the graph of f at x = 5. x^2 y^2 = 3x - 2y. Determine an equation of the tangent line at (2, 1). Use calculus to determine the intervals on which f(x) = x^1 - 5x^4 is concave upward and concave downward. (9 = 6 + 3) Compute the area of the region bounded by the graphs y = x^3 + 2 and y = 2x + 5. Write (you need not evaluate) and integral the will yield the volume of the solid obtained by rotating this region about the x-axis. Write your answers both exactly and as decimals rounded to the nearest hundredth. Consider .12-foot ladder sliding down a wall. Denote by y the Height above ground toe ladder's top, and denote by x the distance from the wall lo the ladder's bottom. Assume the bottom is being pulled away from the wall at a constant rate of 2 ft/sec. Determine the velocity of the top of the ladder when x = 10 feet. A landscaper wishes to enclose a rectangular garden on one side by marble edge costing $60 per foot and on the other three sides by a brick edge costing $20 pet foot. If the area of the garden is 1000 ft^2, determine the dimensions of the garden that minimizes the cost Write your dimensions as decimals accuracy rounded to the nearest hundredth.

Explanation / Answer

5) a) g(x)=6x-x3

differentiate with respect to x .

g'(x)=6-3x2

for critical points g'(x)=0

6-3x2=0

3x2=6

x2=2

x=2

g(1)=6*1-13=5

g(2)=6(2) -(2)3=42

g(3)=6*3 -33

g(3)=-9

absolute maximum =42

absolute minimum =-9

b)[1 to 3] (6x-x3)dx

=[1 to 3] (6(1/2)x2-(1/4)x4)

=[1 to 3] (3x2-(1/4)x4)

=(3*32-(1/4)*34)-(3*12-(1/4)14)

=(27-(81/4))-(3-(1/4))

=4

c)let average value at c

by mean value theorem f(c) *(3-1) =[1 to 3] (6x-x3)dx

average value =(1/(3-1))*[1 to 3] (6x-x3)dx

average value =(1/2)*4

average value =2

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