7. on which interval is the function f(x)=-x³ - x² increasing? x < -1 -1 < x < 0 0 < x < 1 x > 1
7. on which interval is the function f(x)=-x³ - x² increasing? x < -1 -1 < x < 0 0 < x < 1 x > 1
Answer
Explanation:
Step1: Find the derivative
The derivative of $f(x)=-x^{3}-x^{2}$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=-3x^{2}-2x=-x(3x + 2)$.
Step2: Set the derivative greater than 0
We want to find where $f'(x)>0$, so $-x(3x + 2)>0$. This is equivalent to $x(3x + 2)<0$.
Step3: Solve the inequality
The roots of the equation $x(3x + 2)=0$ are $x = 0$ and $x=-\frac{2}{3}$. Using a sign - chart or testing intervals, we find that the solution to $x(3x + 2)<0$ is $-\frac{2}{3}<x<0$. Among the given options, the closest correct interval is $- 1<x<0$.
Answer:
-1 < x < 0