let f(x)=-24/x and g(x)=2-x^3.

Question

1.let f(x)=-24/x and g(x)=2-x^3.

find i (fog)(x), ii (gof)(x) iii (gof)(4), iv (fog)(2)

 

2. a farmer has 600 yards of fencing and wishing to enclose a rectangular garden  whose width is x yards.

 

i Express the area A of the garden as a function of x

ii sketch the graph of A

iii find the values of x for which the area of the garden is largest.

iv Hence determine dimensions of the garden that gives the largest area of the garden.

v find the domain and range of A

 

Answer 1

1. i. (fog)(x) = f(g(x)) = f(2-x³) = -24/(2-x³)

 

    ii. (gof)(x) = g(f(x)) = g(-24/x) = 2-(-24/x)³ = 2+(24/x)³

 

    iii. From ii, (gof)(4) = 2+(24/4)³ = 2+6³ = 218

 

    iv. From i, (fog)(2) = -24/(2-(2)³) = -24/(-6) = 4

 

2. Let x = length and y = width

 

    i. Since the perimeter is 600, 2x + 2y = 600.  So, y = 300 - x

 

       Area = A(x) = (length)(width)

 

       A(x) = x(300-x) = -x²+300x

 

    ii. The graph of y = -x²+300x is a parabola opening downward with           x-intercepts (0,0) and (300,0).  Since x represents the length of           the rectangular garden, x must be positive.  So the graph of                 y = A(x) is the portion of the parabola corresponding to x > 0.

 

    iii. Using the symmetry of the graph, the maximum occurs halfway              between the x-intercepts.  So, the area is largest when                        x = 150 yd.  

 

    iv. Maximum area when x = 150 yd and y = 300-150 = 150 yd.  So,            the garden's area is as large as possible when the garden is a              square with side length 150 yd.                                                        Maximum area = A(150) = (150)(150) = 22,500 yd²

 

     v. A(x) = x(300-x)  

 

         Since x is the length, x > 0

 

         Since y = 300-x is the width, y > 0.  So, 300-x>0.

 

         Therefore, x < 300

 

         Domain = (0, 300)

 

         From the graph of the area function, we see that the range is

         (0, 22,500].