The question in the image reads:

"29. Given $f(x) = 2x^2 - 4x$, determine $f(x) > 0$ and $f^{-1}(x) > 0$. Identify the domain of both $f(x)$ and $f^{-1}(x)$."

Let's break this down step by step.

1. Solve $f(x) > 0$:

We start with the quadratic function $f(x) = 2x^2 - 4x$. First, factor it: $f(x) = 2x(x - 2)$ To solve $f(x) > 0$, we analyze the factored form: $2x(x - 2) > 0$ The critical points are $x = 0$ and $x = 2$. These points divide the number line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

By testing a point in each interval:

• In $(-\infty, 0)$, pick $x = -1$: $2(-1)((-1) - 2) = 2(-1)(-3) = 6 > 0$.
• In $(0, 2)$, pick $x = 1$: $2(1)((1) - 2) = 2(1)(-1) = -2 < 0$.
• In $(2, \infty)$, pick $x = 3$: $2(3)((3) - 2) = 2(3)(1) = 6 > 0$.

Thus, $f(x) > 0$ on the intervals $(-\infty, 0) \cup (2, \infty)$.

2. Solve for $f^{-1}(x)$:

Next, find the inverse of $f(x)$. Start by solving $y = 2x^2 - 4x$ for $x$. Complete the square: $y = 2(x^2 - 2x) = 2((x - 1)^2 - 1) = 2(x - 1)^2 - 2$ Solve for $x$: $y + 2 = 2(x - 1)^2$ $\frac{y + 2}{2} = (x - 1)^2$ $x - 1 = \pm \sqrt{\frac{y + 2}{2}}$ Thus, $x = 1 \pm \sqrt{\frac{y + 2}{2}}$

Since a quadratic function isn't one-to-one without restricting its domain, $f^{-1}(x)$ may not be applicable in the entire range unless restrictions are applied.

3. Domain of $f(x)$:

Since $f(x) = 2x^2 - 4x$ is a quadratic function, its domain is all real numbers, $(-\infty, \infty)$.

4. Domain of $f^{-1}(x)$:

The inverse function $f^{-1}(x)$ will only exist for $x \geq -2$ because of the square root term in its formula. Therefore, the domain of $f^{-1}(x)$ is $[-2, \infty)$.

Would you like any details or further clarifications?

Related Questions:

1. How can we restrict the domain of $f(x)$ to ensure it has an inverse?
2. How can we determine the range of a quadratic function like $f(x)$?
3. What is the graphical interpretation of $f(x) > 0$ and $f^{-1}(x) > 0$?
4. How does completing the square help in finding the inverse of a quadratic function?
5. How can we verify if an inverse function exists for any given function?

Tip: When solving quadratic inequalities, always test points in each interval defined by the critical points.