Find the Local Maxima and Minima f(x)=x^5 natural log of x

Problem

ƒ(x)=x^5*ln(x)

Solution

1. Identify the domain of the function. Since the natural logarithm is only defined for positive values, the domain is x>0

2. Find the first derivative using the product rule, which states (d(u)*v)/d(x)=ud(v)/d(x)+vd(u)/d(x)

d(ƒ(x))/d(x)=x^5d(ln(x))/d(x)+ln(x)d(x^5)/d(x)

d(ƒ(x))/d(x)=x^5*(1/x)+ln(x)*(5*x^4)

d(ƒ(x))/d(x)=x^4+5*x^4*ln(x)

3. Determine the critical points by setting the first derivative equal to zero and solving for x

x^4*(1+5*ln(x))=0

Since x>0 we solve 1+5*ln(x)=0

5*ln(x)=−1

ln(x)=−1/5

x=e^(−1/5)

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

d^2(ƒ(x))/(d(x)^2)=d()/d(x)*(x^4+5*x^4*ln(x))

d^2(ƒ(x))/(d(x)^2)=4*x^3+5*(x^41/x+4*x^3*ln(x))

d^2(ƒ(x))/(d(x)^2)=4*x^3+5*x^3+20*x^3*ln(x)

d^2(ƒ(x))/(d(x)^2)=9*x^3+20*x^3*ln(x)

5. Evaluate the second derivative at the critical point x=e^(−1/5)

ƒ(e^(−1/5))^″=9*(e^(−1/5))^3+20*(e^(−1/5))^3*ln(e^(−1/5))

ƒ(e^(−1/5))^″=9*e^(−3/5)+20*e^(−3/5)*(−1/5)

ƒ(e^(−1/5))^″=9*e^(−3/5)−4*e^(−3/5)

ƒ(e^(−1/5))^″=5*e^(−3/5)

Since 5*e^(−3/5)>0 the function has a local minimum at this point.

6. Calculate the y-coordinate of the local minimum.

ƒ(e^(−1/5))=(e^(−1/5))^5*ln(e^(−1/5))

ƒ(e^(−1/5))=e^(−1)*(−1/5)

ƒ(e^(−1/5))=−1/(5*e)

Final Answer

Local Minimum: *(e^(−1/5),−1/(5*e)), Local Maxima: None

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