Find the Local Maxima and Minima f(x)=x^2 natural log of x

Problem

ƒ(x)=x2*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, where (d(u)*v)/d(x)=ud(v)/d(x)+vd(u)/d(x)

ƒ(x)′=x2d(ln(x))/d(x)+ln(x)d(x2)/d(x)

ƒ(x)′=x21/x+ln(x)*(2*x)

ƒ(x)′=x+2*x*ln(x)

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

x*(1+2*ln(x))=0

Since *x>0,1+2*ln(x)=0

ln(x)=−1/2

x=e(−1/2)

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

ƒ(x)″=d(x+2*x*ln(x))/d(x)

ƒ(x)″=1+2*(x1/x+ln(x)*(1))

ƒ(x)″=1+2*(1+ln(x))

ƒ(x)″=3+2*ln(x)

5. Evaluate the second derivative at the critical point x=e(−1/2)

ƒ(e(−1/2))″=3+2*ln(e(−1/2))

ƒ(e(−1/2))″=3+2*(−1/2)

ƒ(e(−1/2))″=2

Since *ƒ(e(−1/2))″>0, there is a local minimum at *x=e(−1/2).

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

ƒ(e(−1/2))=(e(−1/2))2*ln(e(−1/2))

ƒ(e(−1/2))=e(−1)*(−1/2)

ƒ(e(−1/2))=−1/(2*e)

Final Answer

Local Minimum: *(e(−1/2),−1/(2*e)), Local Maxima: None

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