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

Problem

ƒ(x)=x4−4*x3

Solution

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

d(ƒ(x))/d(x)=4*x3−12*x2

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

4*x3−12*x2=0

4*x2*(x−3)=0

x=0,x=3

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

d2(ƒ(x))/(d(x)2)=12*x2−24*x

4. Evaluate the second derivative at the critical points.

ƒ(0)″=12*(0)2−24*(0)=0

ƒ(3)″=12*(3)2−24*(3)=108−72=36

5. Apply the First Derivative Test for x=0 since the Second Derivative Test was inconclusive (ƒ(0)″=0.

For *x<0,ƒ′*(−1)=4*(−1)3−12*(−1)2=−16<0

For *0<x<3,ƒ(1)′=4*(1)3−12*(1)2=−8<0

Since the sign does not change at *x=0*, it is not a local extremum.

6. Interpret the result for x=3 Since ƒ(3)″>0 the function has a local minimum at x=3

ƒ(3)=(3)4−4*(3)3=81−108=−27

Final Answer

Local Minimum: *(3,−27), Local Maxima: None

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