Case a) f(x)=[x-1]/[x+5] step 1 f(x)=y y=[x-1]/[x+5] step 2 exchange x for y and y for x y=[x-1]/[x+5]------> x=[y-1]/[y+5]----> x*[y+5]=[y-1]----> xy+5x=y-1 step 3 clear the variable y xy+5x=y-1-----> y-xy=5x+1----> y*[1-x]=[5x+1]----> y=[5x+1]/[1-x] step 4 f(x)-1= [5x+1]/[1-x] the function and the inverse function are not the same
case b) g(x)=[x-2]/[x-1] step 1 g(x)=y y=[x-2]/[x-1] step 2 exchange x for y and y for x y=[x-2]/[x-1]------> x=[y-2]/[y-1]----> x*[y-1]=[y-2]----> xy-x=y-2 step 3 clear the variable y xy-x=y-2-----> xy-y=-2+x----> y*[x-1]=[x-2]----> y=[x-2]/[x-1] step 4 g(x)-1= [x-2]/[x-1] the function and the inverse function are the same
case c) h(x)=[x+3]/[x-2] step 1 h(x)=y y=[x+3]/[x-2] step 2 exchange x for y and y for x y=[x+3]/[x-2]------> x=[y+3]/[y-2]----> x*[y-2]=[y+3]----> xy-2x=y+3 step 3 clear the variable y xy-2x=y+3-----> xy-y=3+2x----> y*[x-1]=[2x+3]----> y=[2x+3]/[x-1] step 4 h(x)-1= [2x+3]/[x-1] the function and the inverse function are not the same
case d) k(x)=[x+1]/[x-1] step 1 k(x)=y y=[x+1]/[x-1] step 2 exchange x for y and y for x y=[x+1]/[x-1]------> x=[y+1]/[y-1]---> x*[y-1]=[y+1]----> xy-x=y+1 step 3 clear the variable y xy-x=y+1-----> xy-y=x+1----> y*[x-1]=[x+1]----> y=[x+1]/[x-1] step 4 k(x)-1= [x+1]/[x-1] the function and the inverse function are the same
If x is the length of one side of the square, the total width of the cardboard has two squares cut from it. The amount of width remaining is the width of the box. Since the original width of the cardboard is 10, the width of the box is ...