f(x) = 7 is a even function
Solution:
Given that we have to find the even function
A function is even if and only if f(–x) = f(x)
Steps to follow:
Replace x with -x and compare the result to f(x). If f(-x) = f(x), the function is even.
If f(-x) = - f(x), the function is odd.
If f(-x) ≠ f(x) and f(-x) ≠ -f(x), the function is neither even nor odd.
Option 1
![f(x) = (x - 1)^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/w9f402ntmxn3rd2agseatlxwlb1b9dmva5.png)
Substitute x = -x in above function
![f(-x) = (-x - 1)^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/9i2zlvu7y8duuwv3e6wt5veswdx504fkg0.png)
Thus
![f(-x) \\eq f(x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vfkiw8czicdyhpzah3fos8rhtmli82qtk2.png)
So this is not a even function
Option 2
f(x) = 8x
Substitute x = -x in above function
f(-x) = 8(-x) = -8x
Thus
![f(-x) \\eq f(x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vfkiw8czicdyhpzah3fos8rhtmli82qtk2.png)
So this is not a even function
Option 3
![f(x) = x^2 - x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/6pucyqcgndjb55ussqfb2p37uarwpllj03.png)
Substitute x = -x in above function
![f(-x) = (-x)^2 - (-x) = x^2 + x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8vxmfz6niyrm0a59rcohwk10a1wyleuz62.png)
Thus
![f(-x) \\eq f(x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vfkiw8czicdyhpzah3fos8rhtmli82qtk2.png)
So this is not a even function
Option 4
f(x) = 7
f(-x) = 7
Thus f(-x) = f(x)
Thus it is a even function