Question

given the function ( f(x)=x^{3} ), on what interval of ( x )-values is the graph of ( f(x) ) concave up? (1 point)
( (0, infty) )
( (-infty, 0) )
( (-infty, infty) )
( (0,0) )

Explanation:

Step1: Find the first - derivative

Using the power rule \(y = x^n\), \(y^\prime=nx^{n - 1}\). For \(f(x)=x^{3}\), \(f^\prime(x)=3x^{2}\).

Step2: Find the second - derivative

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

Step3: Determine where \(f^{\prime\prime}(x)>0\)

Set \(f^{\prime\prime}(x)>0\), so \(6x>0\). Divide both sides by 6, we get \(x > 0\).

Answer:

\((0,\infty)\)