Math Problem Statement

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.

Solution

Given the function f(x)=2x2+8x10f(x) = 2x^2 + 8x - 10, let's solve the two parts of the problem:

(1) Finding the least value of kk for which f(x)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)=ax2+bx+cf(x) = ax^2 + bx + c, the x-coordinate of the vertex is given by: x=b2ax = \frac{-b}{2a} For f(x)=2x2+8x10f(x) = 2x^2 + 8x - 10, the coefficients are a=2a = 2, b=8b = 8, and c=10c = -10. Thus, the x-coordinate of the vertex is: x=82(2)=84=2x = \frac{-8}{2(2)} = \frac{-8}{4} = -2 At x=2x = -2, the function reaches its minimum point (since the parabola opens upwards, a=2>0a = 2 > 0).

To make f(x)f(x) one-to-one, we must restrict the domain to either x2x \geq -2 (where the function is increasing) or x2x \leq -2 (where the function is decreasing). The least value of kk that makes f(x)f(x) one-to-one is: k=2k = -2

(2) Expressing f1(x)f^{-1}(x) in terms of xx for the case xkx \geq k:

For the domain x2x \geq -2, we need to find the inverse of the function f(x)=2x2+8x10f(x) = 2x^2 + 8x - 10.

  1. Rewrite the function in vertex form: To make the process easier, we complete the square for f(x)=2x2+8x10f(x) = 2x^2 + 8x - 10.

    Factor out the coefficient of x2x^2 from the first two terms: f(x)=2(x2+4x)10f(x) = 2(x^2 + 4x) - 10

    Complete the square inside the parentheses: x2+4x=(x+2)24x^2 + 4x = (x + 2)^2 - 4 Substituting back: f(x)=2((x+2)24)10=2(x+2)2810=2(x+2)218f(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)y = f(x), so: y=2(x+2)218y = 2(x + 2)^2 - 18 Solving for xx in terms of yy: y+18=2(x+2)2y + 18 = 2(x + 2)^2 y+182=(x+2)2\frac{y + 18}{2} = (x + 2)^2 Taking the square root of both sides: x+2=y+182x + 2 = \sqrt{\frac{y + 18}{2}} Since we are considering the domain where x2x \geq -2, we take the positive square root: x=y+1822x = \sqrt{\frac{y + 18}{2}} - 2

Thus, the inverse function is: f1(x)=x+1822f^{-1}(x) = \sqrt{\frac{x + 18}{2}} - 2

Summary:

  1. The least value of kk for which f(x)f(x) is one-to-one is k=2k = -2.
  2. The inverse function is f1(x)=x+1822f^{-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)=2x2+8x10f(x) = 2x^2 + 8x - 10 for x2x \geq -2?
  3. How can we graph both the function f(x)f(x) and its inverse on the same coordinate plane?
  4. What would happen if we restricted the domain to x2x \leq -2? Would the inverse be different?
  5. How can you algebraically verify that f1(x)f^{-1}(x) is indeed the inverse of f(x)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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Inverse Functions

Formulas

Vertex formula: x = -b / 2a
Completing the square for quadratic equations
Inverse function calculation

Theorems

One-to-one function criteria
Properties of quadratic functions
Inverse function theorem

Suitable Grade Level

Grades 9-12