{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

# Definition 18 suppose f is a function dened on an

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (Why?) Graphically, The graph of h appears to be neither symmetric about the y -axis nor the origin. 4. i(x) = i(−x) = i(−x) = 5x 2x − x3 5(−x) 2(−x) − (−x)3 −5x −2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in fact, equal i(x). However, while this suggests i is even, it doesn’t prove it. (It does, however, 1.7 Graphs of Functions 69 prove i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x) and i(−x) and show they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in i(−x) the numerator is −5x. To re-write i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x) = = = 5x 2x − x3 (−1)5x (−1) (2x − x3 ) −5x −2x + x3 Hence, i(x) = i...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online