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Let f (x) = x4 – 2x3 – 3x2 + 4x + 4, g of x is equal to the square root of the quantity x squared minus x minus 2 end quantity and h of x is equal to the quantity negative x squared plus 1 end quantity over the quantity x squared minus x minus 2 end quantity Part 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)

Sagot :

MrRoyalidn

The domain of f(x) is all set of real values while, the domain of g(x) is x ≤ -1 or x ≥ 2 and both functions have the same range

Part A: Compare the domain and range of the function f(x) to g(x)

The functions are given as:

f(x) = x^4 - 2x^3 - 3x^2 + 4x + 4

g(x) = √(x^2 - x - 2)

Domain

The polynomial function f(x) has no restriction on its input.

So, the domain of f(x) is all set of real values

Set the radical of g(x) = √(x^2 - x - 2) greater than 0

x^2 - x - 2 ≥ 0

Factorize

(x + 1)(x - 2) ≥ 0

Solve for x

x ≥ -1 and x ≥ 2

Combine both inequalities

x ≤ -1 and x ≥ 2

So, the domain of g(x) is x ≤ -1 or x ≥ 2

Range

Using a graphical calculator, we have:

  • Range of f(x) = x^4 - 2x^3 - 3x^2 + 4x + 4 ⇒ f(x) ≥ 0
  • Range of g(x) = √(x^2 - x - 2) ⇒ g(x) ≥ 0

Hence, both functions have the same range

How do the breaks in the domain of h(x) relate to the zeros of f(x)?

We have:

h(x) = (-x^2 + x)/(x^2 - x - 2)

Set the denominator to 0

x^2 - x - 2 = 0

The above represents the radical of the function f(x)

This means that the breaks in the domain of h(x) and the zeros of f(x) are the same

Read more about domain and range at:

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Complete question

Let f(x) = x^4 - 2x^3 - 3x^2 + 4x + 4, g(x) = √(x^2 - x - 2) and h(x) = (-x^2 + x)/(x^2 - x - 2)

Part 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)

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