<span>First we have to determine the slope of each lines by transforming to the slope-intercept form:
y=(3x-7/)4; m2= ¾y=(12x+6)/5, m3 = 12/5
The formula to be used in the proceeding steps is a=tan^-1(m1-m2)/1+m1m2=tan^-1(m1-m2)/1+m1m2
substituting, a=tan^-1(m1-3/4)/1+3m1/4=tan^-1(m1-12/5)1+12m1/5) =>(4m1-3)/(4+3m1)=(5m1-12)/(5+12m1)m1 = -1applying this slope
y -y1 = m(x-x1)
when y1 = 5 and x1 = 4 then,
y - 5 = -1(x-4)
y = -x +4+ 5 ; y = -x +9</span>
Given:
The graph of a function 
.
To find:
The interval where 
.
Solution:
From the given graph graph it is clear that, the function before x=0 and after x=3.6 lies above the x-axis. So, for 
and 
.
The function between x=0 and x=3.6 lies below the x-axis. So, 
for 
.
Now,
For 
, the graph of h(x) is above the x-axis. So, .
For 
, the graph of h(x) is below the x-axis. So, .
For 
, the graph of h(x) is below the x-axis. So, .
Only for the interval , we get .
Therefore, the correct option is A.
