Determine when the graph of the given function f is above the x axis, and when the graph of f is below the x axis. Use interval notation to show where above or below.

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We have the function as

f(x)=(x-1)(x-4)^2(x-7)^2

Put f(x)=0 and then determine the critical value using zero factor property.

(x-1)(x-4)^2(x-7)^2=0

x=1,4,7

So the interval are (-infinity,1)(1,4)(4,7)(7,infinity)

We will considewr each interval and then take any arbitrary value in the interval.If the value of y in the region is less than 0 then it lies below the x-axis and if greater than 0 then it lies above the x-axis.

Consider the interval (-infinity,1)

Take x=0,y=-1*(-4)^2*(-7)^2 =-784

As the value is less than 0 therefore in this interval the y graph lies below the x axis

Consider interval (1,4)

Take x=2

y=(2-1)(2-4)^2(2-7)^2 =100>0

Graph lies above the x-axis

Consider interval (4,7)\Take x=5

y=(5-1)(5-4)^2(5-7)^2 =16>0

Graph lies above the x axis

Consider (7,ininity)

Here also taking x=8,y>0

Graph lies above the x-axis.

So interval below the x-axis is x(- (-infinity,1)

Above the x-axis ,y(- (1,infnity)

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