Let f(x)=sqr. rt.(x^2-6x+9) and g(x)=6-f(x). What is the number of square units in the area of the region enclosed by the graphs of y=f(x) and y=g(x)?

When f(x)=sqr. rt.(x^2-6x+9) and g(x)=6-f(x), the area between graphs of y=f(x) and y=g(x) is 18 square units.

Borys Shumyatskiy | Certified Educator

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It is given that `f ( x ) = sqrt ( x^2 - 6x + 9 ) . ` Let's observe that `x^2 - 6x + 9 = ( x - 3 )^2 , ` so `f ( x ) = | x - 3 | .`

The graph of this function is the graph of `| x | ` shifted 3 units to the right: that is, it consists of two rays which go from the point `( 3 , 0 ) ` into the upper half-plane at angles 45 degrees to the x-axis. The angle between the rays is 90 degrees.

To obtain the graph of `g ( x ) ` from the graph of `f ( x ) , ` reflect this graph over the x-axis and move it 6 units up. This way, the figure below `g ( x ) ` and above `f ( x ) ` is a square with the diagonal of 6 units. Then its side has the length of `6 / sqrt ( 2 ) ` units and its area is `36 / 2 = 18 ` square units.

Of course integration is also a way to find this area. To find the points of intersection of graphs and to integrate, consider `x lt 3 ` where `f ( x ) = 3 - x ` and `x gt= 3 ` where `f ( x ) = x - 3 ` separately.

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