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You should have drawn a graph of the function first. Then, is there a symmetry in the function? What happens if you let ##z = x + \frac 3 2##?songoku said:Let ##f(x)=x(x+1)(x+2)(x+3)=x^4 +6x^3 11x^2 + 6x##
1. Finding the stationary points of ##f(x) \rightarrow f'(x)=0 ##
##4x^3 + 18x^2 + 22x + 6 = 0##
##(2x+3)(4x^2+12x+4)=0##
##x=-\frac{3}{2} , x=\frac{-3 \pm \sqrt{5}}{2}##
Checking the sign diagram of first derivative:
a. For ##x<\frac{-3 - \sqrt{5}}{2} \rightarrow f'(x) = (-)##
b. For ##\frac{-3 - \sqrt{5}}{2}<x<-\frac{3}{2} \rightarrow f'(x) = (+)##
c. For ##-\frac{3}{2} <x<\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (-)##
d. For ##x>\frac{-3 + \sqrt{5}}{2} \rightarrow f'(x) = (+)##
##x=\frac{-3 - \sqrt{5}}{2} \rightarrow y=-1## so ##(\frac{-3 - \sqrt{5}}{2} , -1)## is local minimum
##x=-\frac{3}{2} \rightarrow y=\frac{9}{16}## so ##(-\frac{3}{2} , \frac{9}{16})## is local maximum
##x=\frac{-3 + \sqrt{5}}{2} \rightarrow y=-1## so ##(\frac{-3 + \sqrt{5}}{2} , -1)## is local minimum2. Checking the end behaviour of ##f(x)##
##\lim_{x \to \pm \infty} {f(x)=\infty}##
So the local minimum found above is also global minimum
3.
##f(x)=a## will have no solution if ##a## is below global minimum ##\rightarrow a < -1##
No ##a## will result in ##f(x)## has one solution
##f(x)=a## will have two solutions if ##a## is global minimum or ##a## is above global maximum ##\rightarrow a=-1## or ##a>\frac{9}{16}##
##f(x)=a## will have three solutions if ##a## is global maximum ##\rightarrow a=\frac{9}{16}##
##f(x)=a## will have four solutions if ##a## is between local minimum and local maximum ##\rightarrow -1<a<\frac{9}{16}##
I am not sure how to write good conclusion for final answer
This should give you more insight and save at least some of the algebra.