Which Statement Best Describes The Function F ( X ) = Tan ⁡ ( 3 X + Π 4 ) − 1 F(x)=\tan \left(3 X+\frac{\pi}{4}\right)-1 F ( X ) = Tan ( 3 X + 4 Π ​ ) − 1 ?A. The Function Is Even. B. The Function Is Odd. C. There Is Not Enough Information To Determine Whether The Function Is Even Or Odd. D. The Function

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When dealing with trigonometric functions, it's essential to understand their properties and behavior to determine whether they are even, odd, or neither. In this article, we will explore the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ and determine which statement best describes its function type.

Understanding Even and Odd Functions

Before we dive into the function $f(x)$, let's briefly review what even and odd functions are.

• Even Functions: An even function is a function where $f(-x) = f(x)$ for all x in the domain of the function. This means that if we replace x with -x, the function remains unchanged.
• Odd Functions: An odd function is a function where $f(-x) = -f(x)$ for all x in the domain of the function. This means that if we replace x with -x, the function changes sign but remains unchanged in magnitude.

Analyzing the Function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$

Now that we have a basic understanding of even and odd functions, let's analyze the given function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$.

The function $f(x)$ involves the tangent function, which is a periodic function with a period of $\pi$. The tangent function is also an odd function, as $\tan(-x) = -\tan(x)$.

However, the given function $f(x)$ is not simply the tangent function. It involves a horizontal shift and scaling of the tangent function. The term $3x+\frac{\pi}{4}$ inside the tangent function represents a horizontal shift and scaling of the tangent function.

To determine whether the function $f(x)$ is even, odd, or neither, we need to examine the behavior of the function when we replace x with -x.

Replacing x with -x

Let's replace x with -x in the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$.

$f(-x) = \tan \left(3(-x)+\frac{\pi}{4}\right)-1$ $f(-x) = \tan \left(-3x+\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\left(\tan \left(3 x+\frac{\pi}{4}\right)-1\right)$ $-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

Comparing $f(-x)$ with $-f(x)$, we can see that they are not equal. However, we can simplify $f(-x)$ by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = \tan \left(-3x+\frac{\pi}{4}\right)-1$ $f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

We can see that $f(-x)$ is not equal to $-f(x)$. However, we can simplify $f(-x)$ further by using the property of the tangent function that $\tan(-x) = -\tan(x)$.

$f(-x) = -\tan \left(3x-\frac{\pi}{4}\right)-1$

Now, let's compare $f(-x)$ with $-f(x)$.

$-f(x) = -\tan \left(3 x+\frac{\pi}{4}\right)+1$

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In the previous article, we analyzed the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ and determined that it is not an even or odd function. However, we can still answer some common questions related to this function.

Q: Is the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ periodic?

A: Yes, the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ is periodic. The tangent function has a period of $\pi$, and the term $3x+\frac{\pi}{4}$ inside the tangent function represents a horizontal shift and scaling of the tangent function. Therefore, the function $f(x)$ has a period of $\frac{\pi}{3}$.

Q: Is the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ continuous?

A: Yes, the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ is continuous. The tangent function is continuous everywhere, and the term $3x+\frac{\pi}{4}$ inside the tangent function does not affect the continuity of the function.

Q: Is the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ differentiable?

A: Yes, the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ is differentiable. The tangent function is differentiable everywhere, and the term $3x+\frac{\pi}{4}$ inside the tangent function does not affect the differentiability of the function.

Q: Can we find the derivative of the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$?

A: Yes, we can find the derivative of the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$. Using the chain rule, we can find the derivative of the function as follows:

$f'(x) = 3\sec^2\left(3x+\frac{\pi}{4}\right)$

Q: Can we find the second derivative of the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$?

A: Yes, we can find the second derivative of the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$. Using the chain rule and the product rule, we can find the second derivative of the function as follows:

$f''(x) = 18\sec^2\left(3x+\frac{\pi}{4}\right)\tan\left(3x+\frac{\pi}{4}\right)$

Q: Can we find the third derivative of the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$?

A: Yes, we can find the third derivative of the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$. Using the chain rule, the product rule, and the quotient rule, we can find the third derivative of the function as follows:

$f'''(x) = 54\sec^2\left(3x+\frac{\pi}{4}\right)\tan^2\left(3x+\frac{\pi}{4}\right)+54\sec^4\left(3x+\frac{\pi}{4}\right)$

Conclusion

In this article, we analyzed the function $f(x)=\tan \left(3 x+\frac{\pi}{4}\right)-1$ and answered some common questions related to this function. We determined that the function is periodic, continuous, and differentiable. We also found the first, second, and third derivatives of the function using the chain rule, the product rule, and the quotient rule.

References

• [1] "Trigonometry" by Michael Corral
• [2] "Calculus" by Michael Spivak
• [3] "Differential Equations" by James R. Brannan and William E. Boyce