Find the Local Maxima and Minima 3x^4-16x^3+2016

Problem

3*x4−16*x3+2016

Solution

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

(d(3)*x4−16*x3+2016)/d(x)=12*x3−48*x2

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

12*x3−48*x2=0

12*x2*(x−4)=0

x=0,x=4

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

(d(12)*x3−48*x2)/d(x)=36*x2−96*x

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

36*(4)2−96*(4)=576−384=192

Since 192>0 a local minimum occurs at x=4

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

36*(0)2−96*(0)=0

Since the test is inconclusive, use the First Derivative Test. For x<0 ƒ(x)′<0 For 0<x<4 ƒ(x)′<0 Since the sign does not change, x=0 is an inflection point, not a local extremum.

6. Calculate the function value at the local minimum x=4

3*(4)4−16*(4)3+2016=768−1024+2016=1760

Final Answer

Local Minimum: *(4,1760), Local Maxima: None

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