Find the Local Maxima and Minima f(x)=x^4-50x^2+625

Problem

ƒ(x)=x4−50*x2+625

Solution

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

d(ƒ(x))/d(x)=4*x3−100*x

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

4*x3−100*x=0

4*x*(x2−25)=0

4*x*(x−5)*(x+5)=0

x=0,x=5,x=−5

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

d2(ƒ(x))/(d(x)2)=12*x2−100

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

ƒ(0)″=12*(0)2−100=−100

ƒ(5)″=12*(5)2−100=200

ƒ″*(−5)=12*(−5)2−100=200

5. Interpret the results of the test: ƒ(0)″<0 indicates a local maximum, while ƒ(5)″>0 and ƒ″*(−5)>0 indicate local minima.

Local Maximum at *x=0

Local Minima at *x=5,x=−5

6. Calculate the function values at these points to find the coordinates.

ƒ(0)=0−50*(0)2+625=625

ƒ(5)=5−50*(5)2+625=625−1250+625=0

ƒ*(−5)=(−5)4−50*(−5)2+625=625−1250+625=0

Final Answer

Local Maxima: *(0,625), Local Minima: *(5,0),(−5,0)

Want more problems? Check here!