Math Problem Statement

What is the range of the function f+g(x)=3x2−2+15x2+4f+g(x)=3x2−2+51​x2+4

Solution

To find the range of the function (f+g)(x)=3x22+15x2+4(f + g)(x) = 3x^2 - 2 + \frac{1}{5}x^2 + 4, we need to simplify the expression first and then analyze the resulting function.

Step-by-Step Solution:

  1. Combine the functions:

    Let's simplify the expression for (f+g)(x)(f + g)(x):

    (f+g)(x)=3x22+15x2+4.(f + g)(x) = 3x^2 - 2 + \frac{1}{5}x^2 + 4.

  2. Combine like terms:

    Combine the x2x^2 terms and the constant terms:

    (f+g)(x)=(3+15)x2+(2+4).(f + g)(x) = \left(3 + \frac{1}{5}\right)x^2 + (-2 + 4).

    Let's simplify the coefficients:

    (f+g)(x)=(3+15)x2+2.(f + g)(x) = \left(3 + \frac{1}{5}\right)x^2 + 2.

  3. Convert to a common fraction:

    Convert 33 to a fraction with a denominator of 55:

    3=155.3 = \frac{15}{5}.

    So, we have:

    3+15=155+15=165.3 + \frac{1}{5} = \frac{15}{5} + \frac{1}{5} = \frac{16}{5}.

    Now, rewrite the function:

    (f+g)(x)=165x2+2.(f + g)(x) = \frac{16}{5}x^2 + 2.

  4. Analyze the quadratic function:

    The function (f+g)(x)=165x2+2(f + g)(x) = \frac{16}{5}x^2 + 2 is a quadratic function in the standard form ax2+bx+cax^2 + bx + c, where a=165>0a = \frac{16}{5} > 0, b=0b = 0, and c=2c = 2.

    Since the coefficient of x2x^2 (a=165a = \frac{16}{5}) is positive, the parabola opens upwards.

  5. Determine the range of the function:

    • For a quadratic function that opens upwards, the minimum value occurs at the vertex of the parabola.
    • Since there is no xx term (b=0b = 0), the vertex is at x=0x = 0.

    Plug x=0x = 0 into the function to find the minimum value:

    (f+g)(0)=16502+2=2.(f + g)(0) = \frac{16}{5} \cdot 0^2 + 2 = 2.

    The minimum value of (f+g)(x)(f + g)(x) is 22.

  6. Range of the function:

    • The function can take on any value greater than or equal to 22 because the parabola opens upwards and has a minimum value at 22.

Conclusion:

The range of the function (f+g)(x)=165x2+2(f + g)(x) = \frac{16}{5}x^2 + 2 is:

Range=[2,).\text{Range} = [2, \infty).

Would you like more details, or do you have any questions?

Here are 5 related questions that expand on this information:

  1. How do you find the vertex of a quadratic function in standard form?
  2. What are the effects of the coefficient aa in the quadratic function ax2+bx+cax^2 + bx + c on the graph?
  3. How can you determine the range of a quadratic function that opens downwards?
  4. What transformations occur when you add or subtract a constant to a function?
  5. How do you graph the quadratic function (f+g)(x)=165x2+2(f + g)(x) = \frac{16}{5}x^2 + 2?

Tip: To determine the range of a quadratic function, always look at the direction in which the parabola opens (upwards or downwards) and find the vertex, which represents the minimum or maximum value of the function.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Vertex Form
Range of Quadratic Functions

Formulas

Vertex formula for a quadratic function
Quadratic function standard form

Theorems

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Suitable Grade Level

Grades 10-12