Find the Local Maxima and Minima f(x)=2x^4-8x^3

Problem

ƒ(x)=2*x4−8*x3

Solution

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

d(ƒ(x))/d(x)=8*x3−24*x2

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

8*x3−24*x2=0

8*x2*(x−3)=0

x=0,x=3

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

d2(ƒ(x))/(d(x)2)=24*x2−48*x

4. Test the critical point x=3 by substituting it into the second derivative.

ƒ(3)″=24*(3)2−48*(3)

ƒ(3)″=216−144=72

Since ƒ(3)″>0 there is a local minimum at x=3

5. Test the critical point x=0 by substituting it into the second derivative.

ƒ(0)″=24*(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<3 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=3

ƒ(3)=2*(3)4−8*(3)3

ƒ(3)=162−216=−54

Final Answer

Local Minimum: *(3,−54), Local Maxima: None

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