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

Given the function \( f(x) = 2x^2 + 8x - 10 \), let's solve the two parts of the problem:

### (1) Finding the least value of \( k \) for which \( f(x) \) is one-to-one:

A quadratic function is one-to-one when it is either strictly increasing or strictly decreasing. This happens when we restrict the domain such that the function is monotonic (i.e., either increasing or decreasing on that domain). The key to this is the **vertex** of the quadratic function, which is where the change in direction happens (i.e., where the function stops decreasing and starts increasing or vice versa).

For a quadratic function of the form \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is given by:
\[
x = \frac{-b}{2a}
\]
For \( f(x) = 2x^2 + 8x - 10 \), the coefficients are \( a = 2 \), \( b = 8 \), and \( c = -10 \). Thus, the x-coordinate of the vertex is:
\[
x = \frac{-8}{2(2)} = \frac{-8}{4} = -2
\]
At \( x = -2 \), the function reaches its minimum point (since the parabola opens upwards, \( a = 2 > 0 \)).

To make \( f(x) \) one-to-one, we must restrict the domain to either \( x \geq -2 \) (where the function is increasing) or \( x \leq -2 \) (where the function is decreasing). The least value of \( k \) that makes \( f(x) \) one-to-one is:
\[
k = -2
\]

### (2) Expressing \( f^{-1}(x) \) in terms of \( x \) for the case \( x \geq k \):

For the domain \( x \geq -2 \), we need to find the inverse of the function \( f(x) = 2x^2 + 8x - 10 \).

1. **Rewrite the function in vertex form:**
   To make the process easier, we complete the square for \( f(x) = 2x^2 + 8x - 10 \).

   Factor out the coefficient of \( x^2 \) from the first two terms:
   \[
   f(x) = 2(x^2 + 4x) - 10
   \]

   Complete the square inside the parentheses:
   \[
   x^2 + 4x = (x + 2)^2 - 4
   \]
   Substituting back:
   \[
   f(x) = 2((x + 2)^2 - 4) - 10 = 2(x + 2)^2 - 8 - 10 = 2(x + 2)^2 - 18
   \]

2. **Solve for the inverse:**
   Now, let \( y = f(x) \), so:
   \[
   y = 2(x + 2)^2 - 18
   \]
   Solving for \( x \) in terms of \( y \):
   \[
   y + 18 = 2(x + 2)^2
   \]
   \[
   \frac{y + 18}{2} = (x + 2)^2
   \]
   Taking the square root of both sides:
   \[
   x + 2 = \sqrt{\frac{y + 18}{2}}
   \]
   Since we are considering the domain where \( x \geq -2 \), we take the positive square root:
   \[
   x = \sqrt{\frac{y + 18}{2}} - 2
   \]

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

### Summary:
1. The least value of \( k \) for which \( f(x) \) is one-to-one is \( k = -2 \).
2. The inverse function is \( f^{-1}(x) = \sqrt{\frac{x + 18}{2}} - 2 \).

Do you want further details or have any questions?

Here are 5 related questions:
1. What are the properties of one-to-one functions and how can they be identified?
2. Can you find the range of the function \( f(x) = 2x^2 + 8x - 10 \) for \( x \geq -2 \)?
3. How can we graph both the function \( f(x) \) and its inverse on the same coordinate plane?
4. What would happen if we restricted the domain to \( x \leq -2 \)? Would the inverse be different?
5. How can you algebraically verify that \( f^{-1}(x) \) is indeed the inverse of \( f(x) \)?

**Tip:** Always check the domain and range of a function before finding its inverse, as these may affect the validity of the inverse function.

Explore how to determine the least value of k for which the function f(x) = 2x² + 8x - 10 is one-to-one, and learn how to express its inverse. This problem involves concepts like quadratic functions, vertex formula, and finding inverse functions. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation