If f(x) is an odd function and the graph of f(x) includes points in quadrant IV, which statement about the graph of f(x) must be true?

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Definition: The function f(x) is odd, when f(-x)=f(x) for all x. 
If f(x) is an odd function and the graph of f(x) includes points in quadrant IV, then for these points [tex]x\ge 0, f(x)\le 0[/tex].
When you substitude -x instead of x (-x becomes non-positive), then f(-x) becomes -f(x) and since f(x) is non-positive, then -f(x) is non-negative. Points that have x non-positive and y non-negative lie in the II quadrant.
Conclusion: the graph of f(x) must include points in quadrants II and IV.





The graph of f(x) must include points in quadrants II and IV.

How to determine the true statement?

An odd function f(x) is represented as:

f(-x) = -f(x)

The above implies that:

  • The points in quadrants II are in IV
  • The points in quadrants I are in III

From the question, we understand that graph includes the points in IV.

This means that these points are also in quadrant II

Hence. the graph of f(x) must include points in quadrants II and IV.

Read more about odd functions at:

https://brainly.com/question/14264818

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