Find the Local Maxima and Minima f(x) = natural log of x^4+27

Problem

ƒ(x)=ln(x4+27)

Solution

1. Find the first derivative of the function using the chain rule, where d(ln(u))/d(x)=1/u⋅d(u)/d(x)

d(ƒ(x))/d(x)=1/(x4+27)⋅4*x3

d(ƒ(x))/d(x)=(4*x3)/(x4+27)

2. Identify critical points by setting the first derivative equal to zero and solving for x

(4*x3)/(x4+27)=0

4*x3=0

x=0

3. Apply the First Derivative Test to determine the nature of the critical point by checking the sign of ƒ(x)′ around x=0
   For x<0 (e.g., x=−1:

ƒ′*(−1)=(4*(−1)3)/((−1)4+27)=(−4)/28<0

For x>0 (e.g., x=1:

ƒ(1)′=(4*(1)3)/((1)4+27)=4/28>0

4. Determine the local extrema based on the sign change. Since ƒ(x)′ changes from negative to positive at x=0 there is a local minimum at this point.

ƒ(0)=ln(0+27)

ƒ(0)=ln(27)

Final Answer

Local Minimum: *(0,ln(27)), Local Maxima: None

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