anonymous
  • anonymous
In 1A-4 B, we are trying to show that an even polynomial = even + odd functions. However the starting function on the problem sheet is different than the starting one on the answer sheet. Is this a mistake or an extra step?
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
The equation for f(x) in 1A-4 (b) has a typo. They wanted to write an identity, which we can derive. Begin by writing f(x) this way: \[ f(x)= \frac{f(x)}{2} + \frac{f(x)}{2} \] and then add and subtract f(-x)/2 (which adds zero): \[ f(x)= \frac{f(x)}{2} + \frac{f(x)}{2} + \frac{f(-x)}{2} - \frac{f(-x)}{2} \] and reorder the terms to get \[ f(x)= \frac{f(x)}{2} + \frac{f(-x)}{2}+ \frac{f(x)}{2} - \frac{f(-x)}{2} \\ f(x)= \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} \] if we simplify the right-hand side we get f(x), so the identity is clearly true. But as the answer shows, the two terms represent even and odd functions that add up to create f(x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.