How to graph f ' and justify?
1 Answer
Please see below.
Explanation:
Let us first observe tha function. A few observations are:
- It is always positive.
- Its value is
#0# at#x=+-a# i.e. at#(-a,0)# and#(a,0)#
Hence, it is of the type
At
Before we conclude, let us observe the slope of
-
Initially slope is steeply negative but continues to increase and at
#(-a,0)# it reaches#0# . It continues to increase further till somewhere between#(-a,0)# and#(0,0)# , where we have a point of inflection, where#f''(x)=0# . In other words,#f’’(x)>0# , till point of inflection and while#f’(x)# reaches a local maxima,#f’’(x)=0# at this point between#(-a,0)# and#(0,0)# . -
Beyond this point of inflection, where
#f’’(x))=0# , slope i.e.#f'(x)# falls, till it reaches#0# again at#(0,0)# . It continues to fall, till next point of inflection is reached between#(0,0)# and#(a,0)# . At this point of inflection,#f’(x)# reaches a local minima where#f’(x)<0# and#f’’(x)=0# . -
As we move further right beyond this point of inflection,
#f’(x)# continues to increase reaching#0# at#(a,0)# and beyond this#f’(x)# continues to increase.
As
and hence if
graph{1/4(x^4-6x^2+9) [-5, 5, -1.5, 3.5]}
graph{x(x^2-3) [-10, 10, -5, 5]}