Maths Question HSC 2010 Q9 (1 Viewer)

[cook E]

I need help for this question part (iii)

i) 0≤x<2
ii) 4
iii) -6


Thank you

 

[asianese]

We know that the area above ( from 0 to 2 ) and below (from 2to 4) is equal so that means that the function=0 at this point. Now to calculate f(6), just take the area between x=4 and x=6. It's a rectangle that comes out to be 6u^2. That means f(6) = -6

 

[cook E]

We know that the area above ( from 0 to 2 ) and below (from 2to 4) is equal so that means that the function=0 at this point. Now to calculate f(6), just take the area between x=4 and x=6. It's a rectangle that comes out to be 6u^2. That means f(6) = 0

The answer says f(6)= -6

 

[aphorae]

The answer says f(6)= -6

notice f'(6) = -3
therefore m of f(6) = -3
also you know f(0) is at 0 which means f(4) must be at 0 (it's like a downward parabola with maximum/turning point at 2)

m = rise/run
-3 = x/2
x = -6

 

[barbernator]

f(6)=-6 , as the area under the curve is -6 between 4 and 6.

 

[asianese]

LOL awks sorry typed it wrong. I did mean -6.