Given the f of x equals 2x² plus 8x - 10 for the domain x is greater than or equal to k, part 1 find the least value of k for which f is one to one, part 2 Express f^-1(x) in terms of x in this case.

We are given the function \( f(x) = 2x^2 + 8x - 10 \) and asked to:

1. Find the least value of \( k \) for which \( f \) is one-to-one.
2. Express \( f^{-1}(x) \) (the inverse function) in terms of \( x \), for that domain.

### Part 1: Finding the least value of \( k \)

A function is **one-to-one** (or injective) if it passes the **horizontal line test**, meaning it does not map different values of \( x \) to the same \( y \). For quadratic functions like \( f(x) = 2x^2 + 8x - 10 \), which are parabolas, the function will be one-to-one if we restrict the domain to one side of the vertex (either to the left or right). This is because the parabola either increases or decreases on either side of the vertex, but not both.

1. First, we find the **vertex** of the parabola, as the function is symmetric about this point.
   The \( x \)-coordinate of the vertex of a quadratic function \( f(x) = ax^2 + bx + c \) is given by:
   \[
   x = -\frac{b}{2a}
   \]
   For \( f(x) = 2x^2 + 8x - 10 \), \( a = 2 \) and \( b = 8 \), so:
   \[
   x = -\frac{8}{2(2)} = -2
   \]
   The vertex occurs at \( x = -2 \).

2. Since we need the function to be one-to-one, we can restrict the domain to either \( x \geq -2 \) or \( x \leq -2 \), but the problem asks for the least value of \( k \), so we choose the domain where \( x \geq -2 \). This ensures that the function is increasing and hence one-to-one on this interval.

Therefore, the least value of \( k \) for which \( f(x) \) is one-to-one is \( k = -2 \).

### Part 2: Finding \( f^{-1}(x) \)

Now that we know the domain restriction for \( f(x) \) is \( x \geq -2 \), we can find the inverse function \( f^{-1}(x) \).

1. Set \( f(x) = y \) and solve for \( x \):
   \[
   y = 2x^2 + 8x - 10
   \]
   Rearrange this as a quadratic equation:
   \[
   2x^2 + 8x - (y + 10) = 0
   \]
   This is a quadratic equation in \( x \), so we solve it using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 2 \), \( b = 8 \), and \( c = -(y + 10) \). Substituting these values into the formula:
   \[
   x = \frac{-8 \pm \sqrt{8^2 - 4(2)(-(y + 10))}}{2(2)}
   \]
   Simplifying:
   \[
   x = \frac{-8 \pm \sqrt{64 + 8(y + 10)}}{4}
   \]
   \[
   x = \frac{-8 \pm \sqrt{64 + 8y + 80}}{4}
   \]
   \[
   x = \frac{-8 \pm \sqrt{144 + 8y}}{4}
   \]
   \[
   x = \frac{-8 \pm \sqrt{8(y + 18)}}{4}
   \]
   Simplify further:
   \[
   x = \frac{-8 \pm 2\sqrt{2(y + 18)}}{4}
   \]
   \[
   x = \frac{-4 \pm \sqrt{2(y + 18)}}{2}
   \]
   \[
   x = -2 \pm \frac{\sqrt{2(y + 18)}}{2}
   \]
2. Since we are considering the domain \( x \geq -2 \), we take the positive root:
   \[
   x = -2 + \frac{\sqrt{2(y + 18)}}{2}
   \]
   Now, replace \( y \) with \( x \) to express \( f^{-1}(x) \):
   \[
   f^{-1}(x) = -2 + \frac{\sqrt{2(x + 18)}}{2}
   \]

Thus, the inverse function is:
\[
f^{-1}(x) = -2 + \frac{\sqrt{2(x + 18)}}{2}
\]

---

Would you like more details on any part of this solution or further clarification?

Here are 5 related questions:
1. How do you graph the inverse of a function?
2. Why is the domain restriction necessary for a quadratic function to have an inverse?
3. How do you test if a function is one-to-one using the horizontal line test?
4. What are the properties of quadratic functions and their graphs?
5. How can we verify that a function and its inverse are correct?

**Tip**: When solving for the inverse of a quadratic, always check if you need to restrict the domain, as the original function might not be one-to-one over its entire domain.

Given f(x) = 2x² + 8x - 10 for the domain x ≥ k, part 1: Find the least value of k for which f is one-to-one. Part 2: Express f⁻¹(x) in terms of x in this case.

Learn how to find the least value of k for which f(x) = 2x² + 8x - 10 is one-to-one and how to express its inverse. This problem covers quadratic functions, one-to-one properties, and solving for inverse functions. Suitable for high school students in Grades 10-12.

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