To determine whether the function [tex]\(f(x) = \tan \left(3x + \frac{\pi}{4}\right) - 1\)[/tex] is even, odd, or neither, we must check how it behaves under transformations involving the negative of its argument. Specifically, we will analyze [tex]\(f(-x)\)[/tex]:
1. Define the function [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = \tan(3x + \frac{\pi}{4}) - 1
\][/tex]
2. Find [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = \tan(3(-x) + \frac{\pi}{4}) - 1 = \tan(-3x + \frac{\pi}{4}) - 1
\][/tex]
3. Simplify [tex]\( \tan(-3x + \frac{\pi}{4}) \)[/tex]:
Using the property of the tangent function:
[tex]\[
\tan(-\theta) = -\tan(\theta)
\][/tex]
Therefore,
[tex]\[
\tan(-3x + \frac{\pi}{4}) = -\tan(3x - \frac{\pi}{4})
\][/tex]
4. Substitute back to get [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = -\tan(3x - \frac{\pi}{4}) - 1
\][/tex]
5. Compare [tex]\( f(-x) \)[/tex] and [tex]\( f(x) \)[/tex]:
We now compare [tex]\( f(-x) \)[/tex] to both [tex]\( f(x) \)[/tex] and [tex]\(-f(x)\)[/tex] to determine if the function is even, odd, or neither.
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex].
[tex]\[
f(-x) = -\tan(3x - \frac{\pi}{4}) - 1 \quad \text{and} \quad f(x) = \tan(3x + \frac{\pi}{4}) - 1
\][/tex]
Clearly, [tex]\[f(-x) \neq f(x)\][/tex]
- A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(-x) = -f(x) \)[/tex].
[tex]\[
-f(x) = - (\tan(3x + \frac{\pi}{4}) - 1) = -\tan(3x + \frac{\pi}{4}) + 1
\][/tex]
and,
[tex]\[
f(-x) = -\tan(3x - \frac{\pi}{4}) - 1
\][/tex]
Clearly, [tex]\[f(-x) \neq -f(x)\][/tex]
Since [tex]\( f(-x) \)[/tex] is neither [tex]\( f(x) \)[/tex] nor [tex]\(-f(x)\)[/tex], we can conclude that the function [tex]\( f(x) \)[/tex] is neither even nor odd.
Conclusion:
The correct statement is:
C. The function is neither even nor odd.
