Questions with Solutions

Question 1

Solution to Question 1:

Let us write f(x) as follows
f(x) = (1 / 2) f(x) + (1 / 2) f(x) + (1 / 2) f(-x) - (1 / 2) f(-x)
= (1 / 2) (f(x) + f(-x)) + (1 / 2)(f(x) - f(-x))
Let .
g(x) = (1 / 2) (f(x) + f(-x))
Check that g(x) is even
g(-x) = (1 / 2) (f(-x) + f(x)) = g(x)
Let .
h(x) = (1 / 2) (f(x) - f(-x))
Check that h(x) is odd
h(-x) = (1 / 2) (f(-x) - f(x)) = - (1 / 2) (f(x) - f(-x)) = - h(x)
Hence
f(x) = g(x) + h(x)
where g(x) is even and h(x) is odd and are defined in terms of f(x) above.

Question 2

Solution to Question 2:

f(x) is a polynomial and it is therefore straightforward to separate even and odd parts of the polynomial as follows
f(x) = (2 x⁴ + 2 x² - 4) + (- 5 x³ + x)
where 2 x ⁴ + 2 x ² - 4 is a n even function and -5 x³ + x is an odd function.

Question 3

Solution to Question 3:

In exercise 1, we showed that any function f may be expressed as the sum of an even and an odd functions as follows
f(x) = (1 / 2) (f(x) + f(-x)) + (1 / 2)(f(x) - f(-x))
where g(x) = (1 / 2) (f(x) + f(-x)) is an even function and h(x) = (1 / 2)(f(x) - f(-x)) is an odd function. Hence if f(x) = 1 /(x - 1), then
g(x) = (1 / 2)(1 / (x - 1) + 1 / (- x - 1)) = 1 / (x² - 1)
h(x) = (1 / 2)(1 / (x - 1) - 1 / (- x - 1)) = x / (x² - 1)
We now check our answer. We add g(x) and h(x) and see if the addition gives f(x)
g(x) + h(x) = 1 / (x² - 1) + x / (x² - 1)
= ( 1 + x ) / (x² - 1)
= (1 + x) / [(x - 1)(x + 1)]
= 1 / x - 1 = f(x)

More References on Calculus