Find the Local Maxima and Minima f(x)=x^5-5x^4+100

Problem

ƒ(x)=x^5−5*x^4+100

Solution

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

d(ƒ(x))/d(x)=5*x^4−20*x^3

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

5*x^4−20*x^3=0

5*x^3*(x−4)=0

x=0,x=4

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

d^2(ƒ(x))/(d(x)^2)=20*x^3−60*x^2

4. Test the critical point x=4 by substituting it into the second derivative.

ƒ(4)^″=20*(4)^3−60*(4)^2

ƒ(4)^″=1280−960=320

Since ƒ(4)^″>0 there is a local minimum at x=4

5. Test the critical point x=0 by substituting it into the second derivative.

ƒ(0)^″=20*(0)^3−60*(0)^2=0

Since the Second Derivative Test is inconclusive, use the First Derivative Test. For x<0 ƒ(x)^′>0 For 0<x<4 ƒ(x)^′<0 Since the sign changes from positive to negative, there is a local maximum at x=0

6. Calculate the y-coordinates for the local extrema.

ƒ(0)=0^5−5*(0)^4+100=100

ƒ(4)=4^5−5*(4)^4+100=1024−1280+100=−156

Final Answer

Local Maximum: *(0,100), Local Minimum: *(4,−156)

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