Question

graphing the cubic function quick check
given the function $f(x) = x^3$, on what interval of x-values is the graph of $f(x)$ concave up? (1 po
$(-infty, infty)$
$(0, 0)$
$(0, infty)$
$(-infty, 0)$

Explanation:

Step1: Find the first derivative

To determine concavity, we first find the first derivative of \( f(x) = x^3 \). Using the power rule, the first derivative \( f'(x) = 3x^2 \).

Step2: Find the second derivative

Next, we find the second derivative by differentiating \( f'(x) \). Differentiating \( 3x^2 \) gives \( f''(x) = 6x \).

Step3: Determine where \( f''(x) > 0 \)

To find where the function is concave up, we need to find where the second derivative is positive, i.e., \( f''(x) > 0 \). So we solve \( 6x > 0 \). Dividing both sides by 6 (which is positive, so the inequality direction remains the same), we get \( x > 0 \). In interval notation, this is \( (0, \infty) \).

Answer:

\( (0, \infty) \) (corresponding to the option with this interval)