Prove algebraically if the following functions are even, odd, or neither. (12 points)
1 a. f(x) = x² – 5
b. g(x) = x + 1x
c. h(x) = 5x3 - x
d. f(x) = -3x2 - 5
Show work please :)

  a. even

  b. odd

  c. odd

  d. even

Step-by-step explanation:

We suppose that "prove algebraically" means that you are to substitute -x for x in each equation and show that f(x) = -f(-x) for an odd function, and that f(x) = f(-x) for an even function.

It is easier to do this generically, so that you can make use of the rule that ...

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So, for even-degree, we can assume that the polynomial is of the form ...

  f(x) = ax^2 + b

Then we have ...

  f(-x) = a(-x)^2 +b = (-1)^2·ax^2 +b = ax^2 +b

or ...

  f(-x) = f(x) . . . . regardless of the values of "a" and "b"

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For an odd-degree function, we can assume that the polynomial is of the form ...

  g(x) = x·f(x) . . . . where f(x) is of even degree

Then ...

  g(-x) = -x·f(-x) = -x·f(x) . . . . since f(x) = f(-x) as shown above

or

  g(-x) = -g(x) . . . . . . . . regardless of the coefficients in f(x)

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a. f(x) matches the form of our even-degree f(x) above, so is even.

b. g(x) matches the form of our odd-degree g(x) above (with a=0), so is odd.

c. h(x) = x(5x^2 -1) matches the form of our odd-degree g(x), so is odd.

d. f(x) matches the form of our even-degree f(x) above, so is even.

Comment on the problem

For part (b), we have assumed that "x +1x" means 2x. If it is supposed to be something else, it needs to be properly written. (x+1/x is an odd function, because -x+1/(-x) = -(x +1/x).)

tangent

step-by-step explanation: