Question

basic & cube root functions graphing the cubic function back to intro page click for long desc the equation y equals x superscript 3 baseline plus 2 is written to the right of the curve. approximate coordinates of the curve are as follows: left parenthesis negative 2.2 comma negative 10 right parenthesis, left parenthesis negative 2 comma negative 6 right parenthesis, left parenthesis negative 1 comma 1 right parenthesis, left parenthesis 0 comma 2 right parenthesis, left parenthesis 1 comma 3 right parenthesis, and left parenthesis 2 comma 10 right parenthesis. for what values of x is the cubic function $y = x^3 + 2$ concave down? option #1: $x > 0$ option #2: $x < 0$ option #3: $x = 0$ (1 point)

Explanation:

Step1: Recall Concavity Test

To determine concavity, we use the second derivative. For a function \( y = f(x) \), find the first derivative \( f'(x) \), then the second derivative \( f''(x) \). A function is concave down when \( f''(x) < 0 \).

Step2: Find First Derivative

For \( y = x^3 + 2 \), the first derivative \( y' = \frac{d}{dx}(x^3 + 2) = 3x^2 \) (using power rule: \( \frac{d}{dx}x^n = nx^{n - 1} \), derivative of constant is 0).

Step3: Find Second Derivative

Differentiate \( y' = 3x^2 \) to get the second derivative: \( y'' = \frac{d}{dx}(3x^2) = 6x \) (power rule: \( 3 \times 2x^{2 - 1} = 6x \)).

Step4: Solve \( y'' < 0 \)

We need to find \( x \) such that \( 6x < 0 \). Divide both sides by 6 (positive, so inequality direction remains): \( x < 0 \).

Answer:

Option #2: \( x < 0 \)