using method 1 and 2 to solve | t | > | 2t - 6 |

[aznvolt]

HELP ME SOLVE THISS!!!
it ask somthign about using method 1 and 2..

| t | > | 2t-6 |

 

[arthur ohlsten]

Re: | t | > | 2t-6 |

I t I > I 2t-6 I

1) let us look at y=I t I
y= I t I a V shaped curve open up , vertex at t=0 y=0

2) let us look at
y=I 2t-6I a V shaped curve , open up , vertex at t= 3 y=0

3) sketch both v shaped curves
t>6-2t for 2<t<3
t>2t-6 for 3 <t <6

4) I t I > I 2t-6 I for 2<t<6 ANSWER

please check for errors
Arthur

 

[stapel]

it ask somthign about using method 1 and 2..

What is "method 1 and 2"? (This is something specific to your text, so you will need to provide this information, along with a clear listing of how far you have gotten in applying the required process.)

Thank you!

Eliz.

 

[pka]

I suspect the second method is this: \(\displaystyle \left|t\right|>\left|{2t-6}\right|\quad\Rightarrow\quad t^2>\left( {2t - 6} \right)^2\).

 

[mmm4444bot]

I suspect that the first method is to use the definition of absolute value to get rid of the absolute value signs (i.e., no sketching required).

MY EDIT: corrected ambiguous language

 

[arthur ohlsten]

In my derivation I never said to graph the V shape curves. I suggested you sketch them to visualize the solution
y= ItI
The absolute euation may be replaced by two equations , one for t greater than the vertex value , and one equation below the vertex

y=ItI is
y=t for 0<t
y=-t for t<0

----------------------------------------------------------------
y=I 2t-6 I replace by

y=2t-6 for 3<t
y=6-2t for t<3

-------------------------------------------------------------------------------------

if you sketch, or visualize , the functions you can "see" where ItI > I 2t-6 I
then you can complete the problem

Arthur