Find the Local Maxima and Minima f(x)=x^4-18x^2+81

Problem

ƒ(x)=x4−18*x2+81

Solution

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

d(ƒ(x))/d(x)=4*x3−36*x

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

4*x3−36*x=0

4*x*(x2−9)=0

4*x*(x−3)*(x+3)=0

x=0,x=3,x=−3

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

d2(ƒ(x))/(d(x)2)=12*x2−36

4. Evaluate the second derivative at each critical point to determine if it is a maximum or minimum.

ƒ(0)″=12*(0)2−36=−36

ƒ(3)″=12*(3)2−36=72

ƒ″*(−3)=12*(−3)2−36=72

5. Interpret the results of the Second Derivative Test.
   Since ƒ(0)″<0 there is a local maximum at x=0
   Since ƒ(3)″>0 and ƒ″*(−3)>0 there are local minima at x=3 and x=−3

6. Calculate the y-coordinates by substituting the x values back into the original function ƒ(x)

ƒ(0)=0−18*(0)2+81=81

ƒ(3)=3−18*(3)2+81=81−162+81=0

ƒ*(−3)=(−3)4−18*(−3)2+81=81−162+81=0

Final Answer

Local Maxima: *(0,81), Local Minima: *(3,0),(−3,0)

Want more problems? Check here!