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Let f (x) be a polynomial function with a zero of multiplicity of 1 at 3 and a zero of multiplicity of 2 at 1. Let g(x) be the radical function g of x equals the cube root of x minus 4 Part A: Using the Factor Theorem, determine the polynomial function f (x) in expanded form. Show all necessary calculations. (5 points) Part B: Let h (x) be the piecewise defined function of h of x is the piecewise function of f of x if x is less than 0 and g of x if x is greater than or equal to 0 Are there any breaks in the domain of h (x)? Explain why or why not. (5 points)

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A) The polynomial is:

f(x) = x^3 - 5*x^2 + 7x - 3

B) There are no breaks in the domain of the piecewise function.

How to find the polynomial equation?

We know that f(x) is a polynomial function with the zeros:

  • x = 3, with a multiplicity of 1.
  • x = 1, with a multiplicity of 2.

Then we can write the polynomial f(x) as:

f(x) = a*(x - 3)*(x - 1)*(x - 1)

Where a is the leading coefficient of the polynomial, which is not given in the problem.

Expanding that, we get:

f(x) = a*x^3 - a*5*x^2 + a*7x - a*3

That is the polynomial in the expanded form, where if we take a = 1, we get:

f(x) = x^3 - 5*x^2 + 7x - 3

B) h(x) is a piecewise function, such that the two domains are:

h(x) = f(x)     if x < 0

h(x) = g(x)    if x ≥ 0

Where g(x) = ∛(x - 4)

First, bot functions (polynomial and the cubic root) have the set of all real numbers as their domain, so the domain of the piecewise function is the set of all real numbers.

clearly, we can see that:

f(0) = -3

g(0) =  ∛(- 4)

So we will only have a jump at x = 0, but there are no breaks.

If you want to learn more about polynomials.

https://brainly.com/question/4142886

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