I NEED HELP ON 1-5 PLEASE HELP I REALLY WOULD APPRECIATE IT!!
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32. (a) For an even function, f(x) = f(-x). Given f(5) = 3, we know f(-5) = 3.
Therefore (-5, 3) is also on the graph.
For an odd function, f(-x) = -f(x). Given f(5) = 3, we know f(-5) = -3.
Therefore (-5, -3) is also on the graph.
33. f(-x) = -f(x). The function is odd.
34. f(-x) = x/(x-1) ≠ -f(x) ≠ f(x). The function is neither even nor odd.
35. f(-x) = f(x). The function is even.
Therefore (-5, 3) is also on the graph.
For an odd function, f(-x) = -f(x). Given f(5) = 3, we know f(-5) = -3.
Therefore (-5, -3) is also on the graph.
33. f(-x) = -f(x). The function is odd.
34. f(-x) = x/(x-1) ≠ -f(x) ≠ f(x). The function is neither even nor odd.
35. f(-x) = f(x). The function is even.
Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:
<em>y</em> = -1 - <em>x</em> ==> <em>x</em> = -1 - <em>y</em>
<em>y</em> = <em>x</em> - 1 ==> <em>x</em> = 1 + <em>y</em>
So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.
This means the area is given by the integral,
The integral with respect to <em>x</em> is trivial:
For the remaining integral, integrate term-by-term to get
Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.
