Question

using a graphing calculator, estimate the real zeros, the relative maxima and minima, and the range of the polynomial function.

(g(x) = x^3 - 2.1x + 3)

part 1 of 4

the zero(s) of the function is/are at approximately (x =) box
(round to three decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Identify the function and target

We are given the cubic polynomial function:

$$g(x) = x^3 - 2.1x + 3$$

We need to find the real zero(s) of this function, rounded to three decimal places.

Analyze the behavior of the function

Let's evaluate the function at some integer values to locate the interval containing the real root:

$$g(-2) = (-2)^3 - 2.1(-2) + 3 = -8 + 4.2 + 3 = -0.8$$

$$g(-1) = (-1)^3 - 2.1(-1) + 3 = -1 + 2.1 + 3 = 4.1$$

Since \(g(-2) < 0\) and \(g(-1) > 0\), by the Intermediate Value Theorem, there is a real zero in the interval \((-2, -1)\).

Apply numerical root-finding method

We use Newton's method to find the root precisely. The derivative is:

$$g'(x) = 3x^2 - 2.1$$

Using Newton's iteration formula \(x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}\) with an initial guess \(x_0 = -2\):

$$x_1 = -2 - \frac{g(-2)}{g'(-2)} = -2 - \frac{-0.8}{3(4) - 2.1} = -2 - \frac{-0.8}{9.9} \approx -1.91919$$

$$x_2 = -1.91919 - \frac{g(-1.91919)}{g'(-1.91919)} \approx -1.91919 - \frac{-0.03816}{8.94902} \approx -1.91493$$

$$x_3 = -1.91493 - \frac{g(-1.91493)}{g'(-1.91493)} \approx -1.91493 - \frac{-0.00010}{8.90013} \approx -1.91492$$

Verify the uniqueness of the real root

Let's find the critical points by setting \(g'(x) = 0\):

$$3x^2 - 2.1 = 0 \implies x^2 = 0.7 \implies x \approx \pm 0.837$$

The local maximum is at \(x \approx -0.837\):

$$g(-0.837) \approx (-0.837)^3 - 2.1(-0.837) + 3 \approx -0.586 + 1.758 + 3 = 4.172 > 0$$

The local minimum is at \(x \approx 0.837\):

$$g(0.837) \approx (0.837)^3 - 2.1(0.837) + 3 \approx 0.586 - 1.758 + 3 = 1.828 > 0$$

Since the local minimum value is positive, the function only crosses the x-axis once. Thus, there is only one real zero.

Round to three decimal places

Rounding \(x \approx -1.91492\) to three decimal places gives:

$$x \approx -1.915$$

Answer:

The zero(s) of the function is/are at approximately x = <blank>-1.915</blank>.