Answer The Following For The Function F ( X ) = 2 X 2 E 4 X F(x) = 2x^2 E^{4x} F ( X ) = 2 X 2 E 4 X . (Write 'none' If Any Of The Following Do Not Occur. Give Exact Answers Or At Least 4 Decimal Places).1. $f^{\prime}(x) = $ $\square$2. F F F Is Increasing

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1. $f^{\prime}(x) = $ $\square$

To find the derivative of the function $f(x) = 2x^2 e^{4x}$, we will use the product rule of differentiation. The product rule states that if we have a function of the form $f(x) = u(x)v(x)$, then the derivative of $f(x)$ is given by $f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)$.

In this case, we have $u(x) = 2x^2$ and $v(x) = e^{4x}$. We can find the derivatives of $u(x)$ and $v(x)$ as follows:

• $u^{\prime}(x) = \frac{d}{dx}(2x^2) = 4x$
• $v^{\prime}(x) = \frac{d}{dx}(e^{4x}) = 4e^{4x}$

Now, we can use the product rule to find the derivative of $f(x)$:

$f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)$ $f^{\prime}(x) = (4x)e^{4x} + (2x^2)(4e^{4x})$ $f^{\prime}(x) = 4xe^{4x} + 8x^2e^{4x}$

Combining like terms, we get:

$f^{\prime}(x) = 4xe^{4x} + 8x^2e^{4x}$ $f^{\prime}(x) = e^{4x}(4x + 8x^2)$

2. $f$ is increasing

To determine if the function $f(x) = 2x^2 e^{4x}$ is increasing, we need to find the intervals where the derivative $f^{\prime}(x)$ is positive.

We have found that $f^{\prime}(x) = e^{4x}(4x + 8x^2)$. To find the intervals where $f^{\prime}(x)$ is positive, we need to find the critical points of $f^{\prime}(x)$.

The critical points of $f^{\prime}(x)$ occur when $f^{\prime}(x) = 0$. Setting $f^{\prime}(x) = 0$, we get:

$e^{4x}(4x + 8x^2) = 0$

Since $e^{4x}$ is always positive, we can set $4x + 8x^2 = 0$:

$4x + 8x^2 = 0$ $4x = -8x^2$ $x^2 = -\frac{1}{2}x$ $x^2 + \frac{1}{2}x = 0$ $x(x + \frac{1}{2}) = 0$

Solving for $x$, we get:

$x = 0$ or $x = -\frac{1}{2}$

Now, we need to test the intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 0)$, and $(0, \infty)$ to determine where $f^{\prime}(x)$ is positive.

Let's test a value in each interval:

• For $x = -1$ (in the interval $(-\infty, -\frac{1}{2})$), we have: $f^{\prime}(-1) = e^{-4}(-4 + 8(-1)^2) = e^{-4}(4) > 0$
• For $x = -\frac{1}{4}$ (in the interval $(-\frac{1}{2}, 0)$), we have: $f^{\prime}(-\frac{1}{4}) = e^{-1}(-1 + 8(-\frac{1}{4})^2) = e^{-1}(-1 + \frac{1}{2}) = e^{-1}(-\frac{1}{2}) < 0$
• For $x = 1$ (in the interval $(0, \infty)$), we have: $f^{\prime}(1) = e^{4}(4 + 8(1)^2) = e^{4}(12) > 0$

Since $f^{\prime}(x)$ is positive in the intervals $(-\infty, -\frac{1}{2})$ and $(0, \infty)$, we can conclude that the function $f(x) = 2x^2 e^{4x}$ is increasing in these intervals.

Conclusion

In conclusion, we have found that the derivative of the function $f(x) = 2x^2 e^{4x}$ is given by $f^{\prime}(x) = e^{4x}(4x + 8x^2)$. We have also determined that the function $f(x)$ is increasing in the intervals $(-\infty, -\frac{1}{2})$ and $(0, \infty)$.

Discussion

The function $f(x) = 2x^2 e^{4x}$ is an example of a function that can be used to model real-world phenomena. For instance, the function could be used to model the growth of a population over time, where the population grows at a rate that is proportional to the square of the current population.

The derivative of the function $f(x)$ can be used to determine the rate at which the population is growing. If the derivative is positive, then the population is growing. If the derivative is negative, then the population is declining.

In this case, we have found that the derivative $f^{\prime}(x)$ is positive in the intervals $(-\infty, -\frac{1}{2})$ and $(0, \infty)$. This means that the population is growing in these intervals.

References

• [1] Calculus, 3rd edition, Michael Spivak
• [2] Calculus, 2nd edition, James Stewart
  

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Q: What is the domain of the function $f(x) = 2x^2 e^{4x}$?

A: The domain of the function $f(x) = 2x^2 e^{4x}$ is all real numbers, since the function is defined for all values of $x$.

Q: What is the range of the function $f(x) = 2x^2 e^{4x}$?

A: The range of the function $f(x) = 2x^2 e^{4x}$ is all positive real numbers, since the function is always positive.

Q: Is the function $f(x) = 2x^2 e^{4x}$ continuous?

A: Yes, the function $f(x) = 2x^2 e^{4x}$ is continuous, since it is the product of two continuous functions.

Q: Is the function $f(x) = 2x^2 e^{4x}$ differentiable?

A: Yes, the function $f(x) = 2x^2 e^{4x}$ is differentiable, since it has a derivative.

Q: What is the second derivative of the function $f(x) = 2x^2 e^{4x}$?

A: To find the second derivative of the function $f(x) = 2x^2 e^{4x}$, we need to differentiate the first derivative. Using the product rule and the chain rule, we get:

$f^{\prime\prime}(x) = e^{4x}(4 + 16x) + e^{4x}(8x + 16x^2)$ $f^{\prime\prime}(x) = e^{4x}(4 + 16x + 8x + 16x^2)$ $f^{\prime\prime}(x) = e^{4x}(4 + 24x + 16x^2)$

Q: Is the function $f(x) = 2x^2 e^{4x}$ concave up or concave down?

A: To determine if the function $f(x) = 2x^2 e^{4x}$ is concave up or concave down, we need to examine the second derivative. If the second derivative is positive, then the function is concave up. If the second derivative is negative, then the function is concave down.

We have found that the second derivative is given by $f^{\prime\prime}(x) = e^{4x}(4 + 24x + 16x^2)$. Since the second derivative is always positive, we can conclude that the function $f(x) = 2x^2 e^{4x}$ is concave up.

Q: What is the inflection point of the function $f(x) = 2x^2 e^{4x}$?

A: The inflection point of the function $f(x) = 2x^2 e^{4x}$ is the point where the second derivative changes sign. Since the second derivative is always positive, there is no inflection point.

Q: What is the limit of the function $f(x) = 2x^2 e^{4x}$ as $x$ approaches infinity?

A: To find the limit of the function $f(x) = 2x^2 e^{4x}$ as $x$ approaches infinity, we can use the fact that $e^{4x}$ grows much faster than $2x^2$. Therefore, the limit is given by:

$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} 2x^2 e^{4x} = \infty$

Q: What is the limit of the function $f(x) = 2x^2 e^{4x}$ as $x$ approaches negative infinity?

A: To find the limit of the function $f(x) = 2x^2 e^{4x}$ as $x$ approaches negative infinity, we can use the fact that $e^{4x}$ approaches zero as $x$ approaches negative infinity. Therefore, the limit is given by:

$\lim_{x\to-\infty} f(x) = \lim_{x\to-\infty} 2x^2 e^{4x} = 0$

Q: Is the function $f(x) = 2x^2 e^{4x}$ an even function?

A: Yes, the function $f(x) = 2x^2 e^{4x}$ is an even function, since $f(-x) = f(x)$.

Q: Is the function $f(x) = 2x^2 e^{4x}$ an odd function?

A: No, the function $f(x) = 2x^2 e^{4x}$ is not an odd function, since $f(-x) \neq -f(x)$.

Q: What is the symmetry of the function $f(x) = 2x^2 e^{4x}$?

A: The symmetry of the function $f(x) = 2x^2 e^{4x}$ is even, since $f(-x) = f(x)$.

Q: What is the periodicity of the function $f(x) = 2x^2 e^{4x}$?

A: The periodicity of the function $f(x) = 2x^2 e^{4x}$ is zero, since the function does not repeat itself at regular intervals.

Q: Is the function $f(x) = 2x^2 e^{4x}$ a polynomial function?

A: No, the function $f(x) = 2x^2 e^{4x}$ is not a polynomial function, since it contains an exponential term.

Q: Is the function $f(x) = 2x^2 e^{4x}$ a rational function?

A: No, the function $f(x) = 2x^2 e^{4x}$ is not a rational function, since it contains an exponential term.

Q: Is the function $f(x) = 2x^2 e^{4x}$ a trigonometric function?

A: No, the function $f(x) = 2x^2 e^{4x}$ is not a trigonometric function, since it does not involve trigonometric functions.

Q: Is the function $f(x) = 2x^2 e^{4x}$ a logarithmic function?

A: No, the function $f(x) = 2x^2 e^{4x}$ is not a logarithmic function, since it does not involve logarithmic functions.

Q: Is the function $f(x) = 2x^2 e^{4x}$ an exponential function?

A: Yes, the function $f(x) = 2x^2 e^{4x}$ is an exponential function, since it involves an exponential term.

Q: Is the function $f(x) = 2x^2 e^{4x}$ a power function?

A: Yes, the function $f(x) = 2x^2 e^{4x}$ is a power function, since it involves a power term.

Q: Is the function $f(x) = 2x^2 e^{4x}$ a composite function?

A: Yes, the function $f(x) = 2x^2 e^{4x}$ is a composite function, since it involves the composition of two functions.

Q: What is the inverse of the function $f(x) = 2x^2 e^{4x}$?

A: To find the inverse of the function $f(x) = 2x^2 e^{4x}$, we need to solve the equation $y = 2x^2 e^{4x}$ for $x$. This can be done using the Lambert W function.

Q: What is the derivative of the inverse of the function $f(x) = 2x^2 e^{4x}$?

A: To find the derivative of the inverse of the function $f(x) = 2x^2 e^{4x}$, we need to use the chain rule and the fact that the derivative of the inverse function is given by:

$\frac{d}{dx} f^{-1}(x) = \frac{1}{f^{\prime}(f^{-1}(x))}$

Q: What is the second derivative of the inverse of the function $f(x) = 2x^2 e^{4x}$?

A: To find the second derivative of the inverse of the function $f(x) = 2x^2 e^{4x}$, we need to use the chain rule and the fact that the second derivative of the inverse function is given by:

$\frac{d^2}{dx^2} f^{-1}(x) = \frac{d}{dx} \left( \frac{1}{f^{\prime}(f^{-1}(x))} \right)$

Q: Is the inverse of the function $f(x) = 2x^2 e^{4x}$ an even function?

A: Yes, the inverse of the function $f(x) = 2x^2 e^{4x}$ is an even function, since $f^{-1}(-x) = f^{-1}(x)$.

Q: Is the inverse of the function $f(x) = 2x^2 e^{4x}$