Find the Local Maxima and Minima f(x)=3x^5-20x^3

Problem

ƒ(x)=3*x5−20*x3

Solution

1. Find the first derivative of the function to identify critical points.

d(ƒ(x))/d(x)=15*x4−60*x2

2. Set the derivative to zero and solve for x to find the critical values.

15*x4−60*x2=0

15*x2*(x2−4)=0

15*x2*(x−2)*(x+2)=0

x=0,x=2,x=−2

3. Find the second derivative to apply the Second Derivative Test.

d2(ƒ(x))/(d(x)2)=60*x3−120*x

4. Evaluate the second derivative at each critical point to determine concavity.

ƒ″*(−2)=60*(−2)3−120*(−2)=−480+240=−240

ƒ(0)″=60*(0)3−120*(0)=0

ƒ(2)″=60*(2)3−120*(2)=480−240=240

5. Interpret the results of the Second Derivative Test. Since ƒ″*(−2)<0 there is a local maximum at x=−2 Since ƒ(2)″>0 there is a local minimum at x=2 At x=0 the test is inconclusive, but the First Derivative Test shows the sign of ƒ(x)′ does not change, so it is an inflection point, not a local extremum.

6. Calculate the y-coordinates for the local maxima and minima.

ƒ*(−2)=3*(−2)5−20*(−2)3=−96+160=64

ƒ(2)=3*(2)5−20*(2)3=96−160=−64

Final Answer

Local Maximum: *(−2,64), Local Minimum: *(2,−64)

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