area limited with y = 3/4 - |x - 2|, tangent w/ abscissa x =

[kristina]

Please help me with my problem:

1) Found area of figure, limited with:
a) curve y = 3/4 - |x - 2|
b) tangent in point with abscissa x = 1 and also with straight lines x = 3 and x = 4.

(The expression |x - 2| is absolute value)

The result should be: 3 ln3/2 - 1/6

 

[galactus]

Use parentheses.

Is that:

\(\displaystyle \frac{3}{4-|x-2|}\) or \(\displaystyle \frac{3}{4}-|x-2|\)

 

[kristina]

Use parentheses.

Is that: :

\(\displaystyle \frac{3}{4-|x-2|}\) or \(\displaystyle \frac{3}{4}-|x-2|\)

THE FIRST ONE:
3/(4-Ix-2I)
Please help me!!
Thanks!!

 

[tkhunny]

Still haven't learned to use parentheses, eh? Learn. Why post the same ambiguous question somewhere else? It's still ambiguous. No one is here to make fun of you or to lie to you. If we tell you it needs clarification, then clarify it. Posting the same thing accomplishes nothing.

http://math2.org/mmb/thread/38043

In addition to the ambiguous notation, I suspect no one knows what this means: "tangent in point with abscissa x = 1". Does that mean the line tangent to the curve at x = 1, or (1,1), or y = (4/3) - (x/3)?

Hint: After determining the right limits, you can skip the absolute values. Both 3 and 4 are greater than 2.

 

[kristina]

Hy!!

I` am so sorry! Because my English is so bad. Simply do not know the right phrase for math ,I ` only try to do right and failed.I´was translating word by word , now I´know that this is not ok. I´try again:

1) Find area of figure bounded with :curve y=3/(4-Ix-2I), abscissa x=1, and also with x=3 and x=4.
Result should be 3ln3/2- 1/6