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

Problem

ƒ(x)=3*x5−5*x3

Solution

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

d(ƒ(x))/d(x)=15*x4−15*x2

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

15*x4−15*x2=0

15*x2*(x2−1)=0

15*x2*(x−1)*(x+1)=0

x=0,x=1,x=−1

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

d2(ƒ(x))/(d(x)2)=60*x3−30*x

4. Evaluate the second derivative at each critical point to determine the nature of the extrema.
   For x=−1

ƒ″*(−1)=60*(−1)3−30*(−1)=−60+30=−30

Since ƒ″*(−1)<0 there is a local maximum at x=−1
For x=1

ƒ(1)″=60*(1)3−30*(1)=60−30=30

Since ƒ(1)″>0 there is a local minimum at x=1
For x=0

ƒ(0)″=60*(0)3−30*(0)=0

The Second Derivative Test is inconclusive. Using the First Derivative Test, ƒ(x)′ is negative on both sides of x=0 (e.g., ƒ′*(−0.5)<0 and ƒ(0.5)′<0, so x=0 is an inflection point, not a local extremum.

5. Calculate the y-coordinates for the local maximum and minimum.

ƒ*(−1)=3*(−1)5−5*(−1)3=−3+5=2

ƒ(1)=3*(1)5−5*(1)3=3−5=−2

Final Answer

Local Maximum: *(−1,2), Local Minimum: *(1,−2)

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