Question

question 16 · 1 point consider the function f(x) below. over what open interval(s) is the function increasing and concave up? give your answer in interval notation. enter ∅ if the interval does not exist. f(x)=x^{3}-2x^{2}-21x - 7 provide your answer below:

Explanation:

Step1: Find the first - derivative

First, find the derivative of \(y = f(x)=x^{3}-2x^{2}-21x - 7\) using the power rule \((x^n)^\prime=nx^{n - 1}\).
\(f^\prime(x)=3x^{2}-4x - 21\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(3x^{2}-4x - 21 = 0\). Factor the quadratic equation: \((3x + 7)(x-3)=0\). The critical points are \(x=-\frac{7}{3}\) and \(x = 3\).

Step3: Find the second - derivative

Differentiate \(f^\prime(x)\) to get the second - derivative \(f^{\prime\prime}(x)=6x-4\).

Step4: Determine where \(f^\prime(x)>0\) and \(f^{\prime\prime}(x)>0\)

For \(f^\prime(x)=3x^{2}-4x - 21>0\), the solution of the inequality is \(x<-\frac{7}{3}\) or \(x>3\).
For \(f^{\prime\prime}(x)=6x - 4>0\), the solution is \(x>\frac{2}{3}\).
The intersection of the intervals where \(f^\prime(x)>0\) and \(f^{\prime\prime}(x)>0\) is \(x>3\).

Answer:

\((3,\infty)\)