Which of the following polynomials is guaranteed by the Intermediate Value Theorem to have a zero on the interval [3,4]?

j(x) = 2x² - 5x - 5

General Formulas and Concepts:

Calculus

Step-by-step explanation:

Step 1: Define functions

f(x) = 2x² + x - 3

g(x) = -4x² + 3x - 4

h(x) = x³ - 2x² - x + 4

j(x) = 2x² - 5x - 5

Step 2: Check

Check if each function is continuous over the interval [3, 4].

1. f(x) = 2x² + x - 3

  (a) f is cont on [3, 4]? Yes

2. g(x) = -4x² + 3x - 4

  (a) f is cont on [3, 4]? Yes

3. h(x) = x³ - 2x² - x + 4

  (a) f is cont on [3, 4]? Yes

4. j(x) = 2x² - 5x - 5

  (a) f is cont on [3, 4]? Yes

Step 3: IVT

Find the zeros on the functions in the interval [3, 4].

1. f(x) = 2x² + x - 3

  (b) f(3) = 18 < k = 0 < f(4) = 33? No

2. g(x) = -4x² + 3x - 4

  (b) f(3) = -31 < k = 0 < f(4) = -56? No

3. h(x) = x³ - 2x² - x + 4

  (b) f(3) = 10 < k = 0 < f(4) = 32? No

4. j(x) = 2x² - 5x - 5

  (b) f(3) = -2 < k = 0 < f(4) = 7? Yes

Step 4: Justify IVT Function

j(x) = 2x² - 5x - 5

  (a) f is cont on [3, 4]? Yes

  (b) f(3) = -2 < k = 0 < f(4) = 7? Yes

  ∴ by IVT, there exists a c∈[3, 4] such that f(c) = 0.

step-by-step explanation:

it would open into a square depending on the type of envelope i’m not sure about the second part