question 16 · 1 point consider the function f(x) below. over what open interval(s) is the function…

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:

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:

Answer

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))