Graph x^2 natural log of x

Problem

ƒ(x)=x2*ln(x)

Solution

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

2. Find the x-intercept by setting ƒ(x)=0

x2*ln(x)=0

ln(x)=0⇒x=1

3. Determine the limit as x approaches 0 from the right.

(lim_x→0)(x2)*ln(x)=0

The function approaches 0 but there is a hole at (0,0) because 0 is not in the domain.

4. Find the first derivative to identify critical points using the product rule.

ƒ(x)′=d(x2)/d(x)*ln(x)+x2d(ln(x))/d(x)

ƒ(x)′=2*x*ln(x)+x2*(1/x)

ƒ(x)′=2*x*ln(x)+x

5. Solve for critical points by setting ƒ(x)′=0

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

Since x>0 we solve 2*ln(x)+1=0

ln(x)=−1/2⇒x=e(−1/2)≈0.607

6. Find 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)=−1/(2*e)≈−0.184

7. Find the second derivative to determine concavity and inflection points.

ƒ(x)″=d(2*x*ln(x)+x)/d(x)

ƒ(x)″=2*ln(x)+2*x*(1/x)+1

ƒ(x)″=2*ln(x)+3

8. Solve for inflection points by setting ƒ(x)″=0

2*ln(x)+3=0⇒ln(x)=−3/2

x=e(−3/2)≈0.223

9. Analyze the end behavior. As x→∞ both x2 and ln(x) approach infinity.

(lim_x→∞)(x2)*ln(x)=∞

Final Answer

The graph of ƒ(x)=x2*ln(x) starts with a hole at (0,0) decreases to a local minimum at (e(−1/2),−1/2*e) crosses the x-axis at (1,0) and increases toward infinity as x→∞ It is concave down for 0<x<e(−3/2) and concave up for x>e(−3/2)

Want more problems? Check here!