Find the Local Maxima and Minima f(x)=x^4-12x^3

Problem

ƒ(x)=x4−12*x3

Solution

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

d(ƒ(x))/d(x)=4*x3−36*x2

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

4*x3−36*x2=0

4*x2*(x−9)=0

x=0,x=9

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

d2(ƒ(x))/(d(x)2)=12*x2−72*x

4. Test the critical point x=9 using the second derivative.

ƒ(9)″=12*(9)2−72*(9)

ƒ(9)″=972−648=324

Since ƒ(9)″>0 there is a local minimum at x=9

5. Test the critical point x=0 using the second derivative.

ƒ(0)″=12*(0)2−72*(0)=0

Since the test is inconclusive, use the First Derivative Test. For x<0 ƒ(x)′<0 For 0<x<9 ƒ(x)′<0 Since the sign does not change, x=0 is an inflection point, not a local extremum.

6. Calculate the function value for the local minimum.

ƒ(9)=(9)4−12*(9)3

ƒ(9)=6561−8748=−2187

Final Answer

Local Minimum: *(9,−2187), Local Maxima: None

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