Find the Local Maxima and Minima f(x)=x^4-8x^3

Problem

ƒ(x)=x4−8*x3

Solution

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

(d(x4)−8*x3)/d(x)=4*x3−24*x2

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

4*x3−24*x2=0

4*x2*(x−6)=0

x=0,x=6

3. Find the second derivative to apply the Second Derivative Test for classifying the critical points.

(d(4)*x3−24*x2)/d(x)=12*x2−48*x

4. Evaluate the second derivative at the critical point x=6

ƒ(6)″=12*(6)2−48*(6)

ƒ(6)″=432−288

ƒ(6)″=144

Since ƒ(6)″>0 there is a local minimum at x=6

5. Evaluate the second derivative at the critical point x=0

ƒ(0)″=12*(0)2−48*(0)=0

Since the Second Derivative Test is inconclusive, use the First Derivative Test. For x<0 ƒ(x)′<0 For 0<x<6 ƒ(x)′<0 Since the sign does not change, x=0 is not a local extremum.

6. Calculate the function value at the local minimum x=6

ƒ(6)=(6)4−8*(6)3

ƒ(6)=1296−1728

ƒ(6)=−432

Final Answer

Local Minimum: *(6,−432), Local Maxima: None

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