Description
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.
f(x) = (x-1) (x-4)^2 (x-7)^2
Explanation & Answer
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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)