Find the Local Maxima and Minima f(x)=6x^4+32x^3

Problem

ƒ(x)=6*x4+32*x3

Solution

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

(d(6)*x4+32*x3)/d(x)=24*x3+96*x2

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

24*x3+96*x2=0

24*x2*(x+4)=0

x=0,x=−4

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

(d(24)*x3+96*x2)/d(x)=72*x2+192*x

4. Evaluate the second derivative at the critical points.

ƒ″*(−4)=72*(−4)2+192*(−4)=1152−768=384

ƒ(0)″=72*(0)2+192*(0)=0

5. Interpret the results of the Second Derivative Test. Since ƒ″*(−4)>0 there is a local minimum at x=−4 Since ƒ(0)″=0 the test is inconclusive, so we check the sign of ƒ(x)′ around x=0

ƒ′*(−1)=24*(−1)3+96*(−1)2=72

ƒ(1)′=24*(1)3+96*(1)2=120

Since the sign of ƒ(x)′ does not change at x=0 there is no local extremum at x=0

6. Calculate the function value for the local minimum.

ƒ*(−4)=6*(−4)4+32*(−4)3=1536−2048=−512

Final Answer

Local Minimum: *(−4,−512), Local Maximum: None

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