randRange(1,2) randRange(-4,4) randRange(-4,4) randRange(1,100) a>0?"("+(-2)*a+")":(-2)*a -3*a*p 3*p*(a*a+b*b) -2*a a*a+b*b q===0?"":(q>0?"+"+q+"x^2":"-"+(-q)+"x^2") p===1?"x^3":p+"x^3" cutermf+qutermf+"+"+r+"x+"+s q===0?"":(q>0?"+"+(2*q)+"x":"-"+(-2*q)+"x") (3*p)+"x^2"+lintermfpr+"+"+r q===0?"":(q>0?"+"+(twoqover3p)+"x":"-"+(-twoqover3p)+"x") "x^2"+lintermfprover3p+"+"+rover3p 3*p+"("+alltermsfprover3p+")" -10 10 p*xmin*xmin*xmin+q*xmin*xmin+r*xmin+s p*xmax*xmax*xmax+q*xmax*xmax+r*xmax+s 1000*floor((ymin-20)/1000) 1000*ceil((ymax+20)/1000) [ -10, 10 ] [ flooredymin, ceiledymax ]
This is an interval of decrease problem.
Find the largest (bounded) open interval of decrease for
      f(x)=alltermsf.
If there are no intervals of decrease,
      or if every maximal interval of decrease is unbounded,
      then enter 0 and 0 into the Answer boxes.

0 < x < 0

Differentiating, we obtain:
       f'(x)=alltermsfpr=factoredfpr
Factor f'(x), if possible.

The discriminant of alltermsfprover3p is
       minustwoa^2-4(1)((a*a+b*b)) =4*a*a-4*(a*a+b*b),
   which is negative, so f'(x) does not factor (over the real numbers),
   so f'(x) has no (real) roots.

The derivative f'(x) is positive for large positive numbers x.

Thus, the derivative f'(x) is positive everywhere.

Then the function f(x) is increasing everywhere,
   and so f(x) has no intervals of decrease.

Following the directions in the problem, enter 0 and 0 into the Answer boxes,    to indicate that there are no intervals of decrease.

NOTE: The next two hints show: the graph of f, followed by the graph of f'.
Note that the graph of f always runs uphill.
Note that the graph of f' is always above the x-axis.

initAutoscaledGraph( [ xrange , yrange ], {} ); style({ stroke: "#6495ED", strokeWidth: 3 }, function() { plot( function( x ) { return p*x*x*x+q*x*x+r*x+s; }, [-10,10] ); });

initAutoscaledGraph( [ [ -10, 10] , [ -500, 500] ], {} ); style({ stroke: "#6495ED", strokeWidth: 3 }, function() { plot( function( x ) { return 3*p*x*x+2*q*x+r; }, [-10,10] ); });