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Is f(x) =(x+5)2 a function, an odd function, both or neither

2 Answers

3 votes

f(x) =(x+5)² is indeed a function. This is because, for every value of 'x' that we chose, there is exactly one corresponding 'y' that can be calculated using the formula (x+5)². A function must have this property in order to be a function - that each 'x' value can be mapped to exactly one 'y' value.

Let's now check if it's an odd function. For a function to be odd, it must have the property that for all 'x', the function of '-x' is equal to the negative of the function of 'x'. That is, f(-x) = -f(x) for all 'x'.

If we apply '-x' into our function, we have (-x + 5)². Now, if we calculate the negative of the function of 'x', -f(x), we have -((x + 5)²).

By comparing f(-x) and -f(x), we can see that they are not equal. Therefore, f(x) =(x+5)² does not have the required property to be an odd function.

So our function f(x) =(x+5)² is a function, but not an odd function.

User ArturM
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2 votes

Answer: Neither even, nor odd

Step-by-step explanation:

f(x) = (x+5)^2 = x^2+10x+25 = x^2+10x^1+25x^0

The exponents for that last expression are 2, 1, 0

The mix of even and odd exponents in the standard form means f(x) is neither even nor odd. We would need to have all exponents even to have f(x) even, or have all exponents odd to have f(x) be odd.

User Roman Khimov
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5.1k points