Question

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)
the best answer is option #

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y=x^{3}+2\), \(y^\prime=\frac{d}{dx}(x^{3}+2)=3x^{2}\)

Step2: Find the second derivative

Differentiate \(y^\prime = 3x^{2}\) with respect to \(x\). Using the power rule again, \(y^{\prime\prime}=\frac{d}{dx}(3x^{2}) = 6x\)

Step3: Determine concavity

A function \(y = f(x)\) is concave - down when \(y^{\prime\prime}<0\). Set \(y^{\prime\prime}=6x<0\). Solving the inequality \(6x < 0\) gives \(x<0\)

Answer:

Option #2: \(x < 0\)