How do you determine if #f(x) = 0.7x^2 + 3# is an even or odd function?

Question

Chloe Alexander · Accepted Answer

Even function

We have: #f(x) = 0.7 x^(2) + 3# For a function, if #f(x) = f(- x)#, then it is said to be even. #=> f(- x) = 0.7 (- x)^(2) + 3# #=> f(- x) = 0.7 x^(2) + 3# Therefore, the function is even.

Cora Alexander · Answer

undefined To determine if a function is even or odd, we can apply the following tests:

1. Even Function Test: A function f(x) is even if f(x) = f(-x) for all x in its domain.

2. Odd Function Test: A function f(x) is odd if f(x) = -f(-x) for all x in its domain.

Let's apply these tests to the function f(x) = 0.7x^2 + 3:

1. Even Function Test:
   f(x) = 0.7x^2 + 3
   f(-x) = 0.7(-x)^2 + 3 = 0.7x^2 + 3

Since f(x) = f(-x), the function f(x) = 0.7x^2 + 3 is even.

2. Odd Function Test:
   f(x) = 0.7x^2 + 3
   -f(-x) = -(0.7(-x)^2 + 3) = -(0.7x^2 + 3) = -0.7x^2 - 3

Since f(x) ≠ -f(-x), the function f(x) = 0.7x^2 + 3 is not odd.

Therefore, the function f(x) = 0.7x^2 + 3 is even.