Question

Let f (x) = x4 + 4x3 – 2x2 – 12x + 9, g of x is equal to the square root of the quantity x squared plus 2 times x minus 3 end quantity and h of x is equal to the quantity negative x squared plus 1 end quantity over the quantity x squared plus 2 times x minus 3 end quantityPart A: Use complete sentences to compare the domain and range of the polynomial function f (x) to that of the radical function g(x). (5 points)Part B: How do the breaks in the domain of h (x) relate to the zeros of f (x)? (5 points)

Answer 1

The zeros of a function are the values of x when the function, f(x) is 0

The breaks in a function is a point where the function is discontinuous

The, domain, breaks and relationships of the functions f(x), g(x), and h(x) are as follows;

Part A:

The domain and range of the functions f(x) and g(x) are D(-∞, ∞), R(-∞, ∞) and D(-∞, -3] $\bigcup$ [1, ∞), R[0, ∞) respectively

Part B:

The values of the breaks of the function h(x) and the zeros of the function f(x) are equal

The reason the above values are correct are as follows:

The given parameters are;

f(x) = x⁴ + 4·x³ - 2·x² - 12·x + 9

$\mathbf{g(x)} = \sqrt{x^2 + 2\times x-3}$

$\mathbf{h(x)} = \dfrac{-x^2 + 1}{x^2+ 2 \times x- 3}$

Part A: The function, f(x) = x⁴ + 4·x³ - 2·x² - 12·x + 9, has a value for all input, value, x, therefore, the domain of the function is D: -∞ < x < ∞

The range of the function f(x) is R: -∞ < y < ∞

The value of x² + 2·x -3 is less than 0 for -3 < x < 1

Therefore, the function, g(x) is not defined for -3 < x < 1, and the domain of the function is given as follows; -∞ < x ≤-3 $\bigcup$ 1 ≤ x < ∞

The range of the function g(x) is R: 0 ≤ y < ∞

Therefore, the comparison is as follows;

The domain of the function f(x) is continuous and is D(-∞, ∞) . The domain of the function g(x) has a discontinuity from x = -3 to x = 1, and is therefore D(-∞, -3] $\bigcup$ [1, ∞)

The range of the function is R(-∞, ∞), while the range of g(x) is R[0, ∞)

The functions intersect and have equal values at (-3, 0), and (1, 0)

Part B:

The breaks of the function h(x) is given by the zeros of the denominator, of h(x)

The zeros of the denominator of h(x), x² + 2·x - 3 are; x = -3, and x = 1

The zeros of f(x) = x4 + 4x3 – 2x2 – 12x + 9, are; x = -3, and x = 1

Therefore, the breaks of the function h(x) are at x = -3, and x = 1, which are the zeros of the function f(x)

The breaks of the function h(x) are at the same location and are equal to the zeros of the function f(x)

Learn more about the radical function and the breaks in the domain of a function here:

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