he graph of the function has one relative maximum and relative minimum point.

Question

the graph of the function has one relative maximum and relative minimum point. plot these two points and check the concavity there. Using only this information, sketch the graph. F(X)=x^3-12x

The relative minimum point on the graph is______(type an ordered pair.)

since the value of F " concave up, concave down , changing concave . at this relative minimum point is ____at  this point , the graph is  

the relative maximum point on the graph is?____(type an ordered pair.)

since the value of F " concave up, concave down , changing concave . at this relative maximum point is ____at  this point , the graph is

Answer 1

f(x)=x^3-12x=x(x^2-12), and f'(x)=3x^2-12=0 at extrema: x^2=4 so x=+2. f(2)=-16, f(-2)=16, so the points are (2,-16) and (-2,16). The graph passes through zero, implying that one turning point is maximum and the other minimum. f(1)=-11 and f(-1)=11, so we can deduce that (2,-16) is a minimum and (-2,16) is a maximum because f(x) in the vicinity has a value nearer to zero then the turning point. f"(x)=6x, f"(2)>0 (minimum), f"(-2)<0 (maximum).