f(x)=alltermsf
.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 negative for large positive numbers x
.
Thus, the derivative f'(x)
is negative everywhere.
Then the function f(x)
is decreasing everywhere,
and so f(x)
has no intervals of increase.
Following the directions in the problem, enter 0
and 0
into the Answer boxes,
to indicate that there are no intervals of increase.
NOTE: The next two hints show: the graph of f
, followed by the graph of f'
.
Note that the graph of f
always runs downhill.
Note that the graph of f'
is always below 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] ); });