- #71
songoku
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- 351
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
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
I thought if I drew the graph, no algebra needed anymore because I could just draw horizontal lines and counted the number of intersections?PeroK said:You should have drawn a graph of the function first. Then, is there a symmetry in the function?
Sorry I don't understand this part. You mean I equate ##f(x)## with ##z=x+ \frac 3 2##?What happens if you let ##z = x + \frac 3 2##?
fresh_42 said:Can you summarize this to check the answers which are somehow hidden in your text, which is hard to follow, since you didn't say what you are doing there?
No solution: ##a = ... ## or ## a\in ...##
1 solution: ##a = ... ## or ## a\in ...##
2 solutions: ##a = ... ## or ## a\in ...##
etc.
My answer is in part (3) of my working.
a) no solution ##\rightarrow a<-1##
b) 1 solution ##\rightarrow a=\emptyset##
c) 2 solutions ##\rightarrow a=-1 ~\text{or}~a>\frac{9}{16}##
d) 3 solutions ##\rightarrow a=\frac{9}{16}##
e) 4 solutions ##\rightarrow -1<a<\frac{9}{16}##
f) more than 4 solutions ##\rightarrow a=\emptyset##
I need to think about thisI know I haven't explicitly asked for, but what are the solutions? At which values of ##x## does the graph intersect the ##x-##axis in the five cases?
I will edit my post to include thisfresh_42 said:... and ##14b = 458752##. This would have actually answered the question!
Edit: sorry can not edit post#49 anymore
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