Honors Even and Odd Functions wkst - Free Printable

Let's solve each part of the problem step by step, based on the definitions provided:

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Definitions Recap:

- Even function: Symmetric about the y-axis → $ f(-x) = f(x) $
- Odd function: Symmetric about the origin → $ f(-x) = -f(x) $
- A function can be neither even nor odd if it doesn't satisfy either condition.

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## 1. Indicate which of the following functions are even, which are odd, and which are neither.

We analyze the graphs:

Graph (a):

- This graph is a V-shape opening downward, symmetric about the y-axis.
- It looks like $ f(x) = -|x| $.
- Symmetric about y-axis → Even function

✔ Even

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Graph (b):

- This is a parabola opening downward, vertex at origin.
- Symmetric about the y-axis.
- Example: $ f(x) = -x^2 $
- Symmetric about y-axis → Even function

✔ Even

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Graph (c):

- This is a straight line passing through the origin with positive slope.
- The graph passes through the origin and has symmetry about the origin.
- For example, $ f(x) = x $
- $ f(-x) = -f(x) $ → Odd function

✔ Odd

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Graph (d):

- This is a W-shaped graph, symmetric about the y-axis.
- Looks like $ f(x) = |x| $ or $ f(x) = x^4 $, but in this case, it’s symmetric about the y-axis.
- So, $ f(-x) = f(x) $

✔ Even

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✔ Summary for Question 1:

- (a): Even
- (b): Even
- (c): Odd
- (d): Even

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## 2. Algebraically determine whether each function is odd, even, or neither.

We test:
- $ f(-x) = f(x) $ → Even
- $ f(-x) = -f(x) $ → Odd
- Neither → If neither condition holds

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a) $ f(x) = 3x^5 - 5x^3 + 17 $

Compute $ f(-x) $:
$$
f(-x) = 3(-x)^5 - 5(-x)^3 + 17 = 3(-x^5) - 5(-x^3) + 17 = -3x^5 + 5x^3 + 17
$$

Compare to $ f(x) $ and $ -f(x) $:

- $ f(x) = 3x^5 - 5x^3 + 17 $
- $ -f(x) = -3x^5 + 5x^3 - 17 $

But $ f(-x) = -3x^5 + 5x^3 + 17 \neq f(x) $, and $ \neq -f(x) $

So, neither.

✘ Neither

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b) $ f(x) = |x| $

$ f(-x) = |-x| = |x| = f(x) $

→ $ f(-x) = f(x) $

✔ Even

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c) $ f(x) = 12x^3 + 6x^2 - 2x $

Compute $ f(-x) $:
$$
f(-x) = 12(-x)^3 + 6(-x)^2 - 2(-x) = -12x^3 + 6x^2 + 2x
$$

Now compare:
- $ f(x) = 12x^3 + 6x^2 - 2x $
- $ -f(x) = -12x^3 - 6x^2 + 2x $

Is $ f(-x) = f(x) $? No.
Is $ f(-x) = -f(x) $? Let's check:

$ f(-x) = -12x^3 + 6x^2 + 2x $
$ -f(x) = -12x^3 - 6x^2 + 2x $

They differ in the $ x^2 $ term → Not equal.

So, neither.

✘ Neither

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d) $ f(x) = 4x^3 - 7 $

$ f(-x) = 4(-x)^3 - 7 = -4x^3 - 7 $

Compare:
- $ f(x) = 4x^3 - 7 $
- $ -f(x) = -4x^3 + 7 $

$ f(-x) = -4x^3 - 7 \neq f(x) $, and $ \neq -f(x) $

So, neither

✘ Neither

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e) $ f(x) = x^2 + 2x + 2 $

$ f(-x) = (-x)^2 + 2(-x) + 2 = x^2 - 2x + 2 $

Compare:
- $ f(x) = x^2 + 2x + 2 $
- $ -f(x) = -x^2 - 2x - 2 $

$ f(-x) \neq f(x) $, $ f(-x) \neq -f(x) $

✘ Neither

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f) $ f(x) = \frac{x^2 - 5}{2x^2 + x} $

Compute $ f(-x) $:
$$
f(-x) = \frac{(-x)^2 - 5}{2(-x)^2 + (-x)} = \frac{x^2 - 5}{2x^2 - x}
$$

Compare to $ f(x) = \frac{x^2 - 5}{2x^2 + x} $

Clearly, $ f(-x) \neq f(x) $, because denominator changes sign in the linear term.

Check $ -f(x) = -\frac{x^2 - 5}{2x^2 + x} $

But $ f(-x) = \frac{x^2 - 5}{2x^2 - x} \neq -f(x) $

So, not even, not odd.

✘ Neither

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✔ Summary for Question 2:

| Function | Type |
|--------|------|
| a) $ 3x^5 - 5x^3 + 17 $ | Neither |
| b) $ |x| $ | Even |
| c) $ 12x^3 + 6x^2 - 2x $ | Neither |
| d) $ 4x^3 - 7 $ | Neither |
| e) $ x^2 + 2x + 2 $ | Neither |
| f) $ \frac{x^2 - 5}{2x^2 + x} $ | Neither |

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## 3. The graphs of an odd function are symmetric about the origin. What geometric property characterizes even functions?

Even functions are symmetric about the y-axis.

That means if you reflect the graph across the y-axis, it remains unchanged.

✔ Answer: Even functions are symmetric about the y-axis.

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## 4. Suppose $ f $ is an odd function whose domain includes zero. Explain why $ f(0) = 0 $ must be true.

Since $ f $ is odd, we have:
$$
f(-x) = -f(x) \quad \text{for all } x \text{ in domain}
$$

Let $ x = 0 $. Then:
$$
f(-0) = -f(0)
\Rightarrow f(0) = -f(0)
$$

Add $ f(0) $ to both sides:
$$
f(0) + f(0) = 0 \Rightarrow 2f(0) = 0 \Rightarrow f(0) = 0
$$

✔ Therefore, if $ f $ is odd and defined at 0, then $ f(0) = 0 $.

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## 5. Can a function be both even and odd? (Hint: Consider constant functions of the form $ f(x) = c $)

Let’s suppose $ f $ is both even and odd.

Then:
- $ f(-x) = f(x) $ → even
- $ f(-x) = -f(x) $ → odd

Set them equal:
$$
f(x) = -f(x) \Rightarrow 2f(x) = 0 \Rightarrow f(x) = 0
$$

So the only function that satisfies both conditions is the zero function:
$$
f(x) = 0 \quad \text{for all } x
$$

This is indeed both even and odd:
- $ f(-x) = 0 = f(x) $ → even
- $ f(-x) = 0 = -0 = -f(x) $ → odd

✔ Yes, a function can be both even and odd — only if it is identically zero.

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✔ Final Answer to Question 5:

Yes, a function can be both even and odd only if it is the zero function, i.e., $ f(x) = 0 $ for all $ x $.

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🟩 Final Answers Summary:

#### 1. Graphs:
- (a): Even
- (b): Even
- (c): Odd
- (d): Even

#### 2. Algebraic Functions:
- a) Neither
- b) Even
- c) Neither
- d) Neither
- e) Neither
- f) Neither

#### 3. Geometric Property of Even Functions:
Symmetric about the y-axis

#### 4. Why $ f(0) = 0 $ for odd functions with domain including 0:
Because $ f(0) = -f(0) \Rightarrow f(0) = 0 $

#### 5. Can a function be both even and odd?
Yes, only if $ f(x) = 0 $ for all $ x $. This function is both even and odd.

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