Answer to Question #241991 in Calculus for asdasdas

Question

Given the function f(x)=-x^3/3+x^2-6x-2, discuss its relative maximum and minimum

points, the intervals where it is increasing and decreasing, the intervals of concavity, and the points of inflection. Construct a sketch of the graph of the function.

Answer 1

Given the function $f(x)=-\frac{x^3}3+x^2-6x-2$, let us discuss its relative maximum and minimum

points, the intervals where it is increasing and decreasing, the intervals of concavity, and the points of inflection.

Since $f'(x)=-x^2+2x-6=-(x^2-2x+1)-5=-(x-1)^2-5<0,$ we conclude that the function is decreasing on the set $\R$ of real numbers. It has no relative maximum and minimum points.

Taking into account that $f''(x)=-2x+2,$ and $f''(x)=0$ implies $x=1,$ we conclude that $x=1$ is the point of inflection. Since $f''(x)>0$ for $x<1,$ we get that on the interval $(-\infty,1)$ the function is concave up. Since $f''(x)<0$ for $x>1,$ we conclude that on the interval $(1,+\infty)$ the function is concave down.

Let us construct a sketch of the graph of the function: