Math Problem Statement

The user submitted an image containing two math problems, one related to finding the zeros, range, and specific values of a function, and the other involving intersections and zeros of quadratic functions.

Solution

Let's break down the tasks based on the image you provided, which consists of two exercises: 4.9 and 4.10.

4.9

This question refers to a graph of a function ff with the domain Df=[1,5]D_f = [-1, 5].

a. Finn nullpunktene til ff

Find the zeros of the function ff (the points where the graph intersects the x-axis).

  • From the graph, the function crosses the x-axis at x=0x = 0 and x=4x = 4.
  • Therefore, the zeros of ff are x=0x = 0 and x=4x = 4.

b. Hva er verdimengden til ff?

Find the range of the function ff.

  • From the graph, the minimum value of f(x)f(x) is 4-4 (at x3x \approx 3) and the maximum value is 1010 (at x=1x = -1).
  • Thus, the range of ff is [4,10][-4, 10].

c. Finn f(0)f(0)

Find the value of f(0)f(0).

  • Looking at the graph, when x=0x = 0, f(x)=4f(x) = 4.
  • Therefore, f(0)=4f(0) = 4.

d. For hvilke verdier av xx er f(x)=4f(x) = 4?

Find the values of xx for which f(x)=4f(x) = 4.

  • From the graph, f(x)=4f(x) = 4 at x=0x = 0 and x=2x = 2.
  • Therefore, the values of xx where f(x)=4f(x) = 4 are x=0x = 0 and x=2x = 2.

4.10

The functions f(x)=x24f(x) = x^2 - 4 and g(x)=4x2g(x) = 4 - x^2 are given.

a. Finn eventuelle toppunkter og bunnpunkter på grafene

Find the possible maximum and minimum points on the graphs.

  • For f(x)=x24f(x) = x^2 - 4:

    • This is a parabola opening upwards. The vertex, which is the minimum point, is at x=0x = 0, and f(0)=4f(0) = -4. Therefore, the minimum point is (0,4)(0, -4).
  • For g(x)=4x2g(x) = 4 - x^2:

    • This is a parabola opening downwards. The vertex, which is the maximum point, is at x=0x = 0, and g(0)=4g(0) = 4. Therefore, the maximum point is (0,4)(0, 4).

b. Finn skjæringspunktene mellom grafene

Find the intersection points of the graphs.

  • To find where f(x)=g(x)f(x) = g(x), solve x24=4x2x^2 - 4 = 4 - x^2. 2x2=8x2=4x=±22x^2 = 8 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2
    • For x=2x = 2, f(2)=g(2)=0f(2) = g(2) = 0, and for x=2x = -2, f(2)=g(2)=0f(-2) = g(-2) = 0.
    • Therefore, the intersection points are (2,0)(2, 0) and (2,0)(-2, 0).

c. Finn nullpunktene til funksjonene

Find the zeros of the functions.

  • For f(x)=x24f(x) = x^2 - 4, solve x24=0x^2 - 4 = 0. The solutions are x=±2x = \pm 2.

    • The zeros of f(x)f(x) are x=2x = 2 and x=2x = -2.
  • For g(x)=4x2g(x) = 4 - x^2, solve 4x2=04 - x^2 = 0. The solutions are also x=±2x = \pm 2.

    • The zeros of g(x)g(x) are x=2x = 2 and x=2x = -2.

Do you need more details or clarifications on any of these points?
Here are some additional questions that might expand on the above content:

  1. What are the exact coordinates of the minimum and maximum points for other quadratic functions?
  2. How can you verify the intersection points of two functions algebraically?
  3. Can you explore the vertex form of a quadratic equation to analyze the function behavior?
  4. What is the significance of the range of a function in real-world applications?
  5. How do the zeros of a function relate to its graphical representation?

Tip: Remember that the zeros of a function are the points where the graph intersects the x-axis. These points give valuable information about the behavior of the function.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Zeros of Functions
Range of Functions
Function Intersections

Formulas

Quadratic equation f(x) = ax^2 + bx + c
Vertex form of a quadratic function
Equation of a parabola: f(x) = x^2 - 4 and g(x) = 4 - x^2

Theorems

Quadratic Formula
Parabola Properties
Function Intersection and Zeros

Suitable Grade Level

Grades 8-10