Graph f(x) = natural log of x^4+27

Problem

ƒ(x)=ln(x4+27)

Solution

1. Identify the domain of the function. Since the argument of a natural logarithm must be positive, we require x4+27>0 Because x4≥0 for all real x the expression x4+27 is always at least 27 Thus, the domain is all real numbers, (−∞,∞)

2. Determine the symmetry of the function. We evaluate ƒ*(−x)=ln((−x)4+27)=ln(x4+27) Since ƒ*(−x)=ƒ(x) the function is even and symmetric about the yaxis.

3. Find the intercepts. To find the yintercept, evaluate ƒ(0)=ln(0+27)=ln(27)≈3.3 To find xintercepts, set ƒ(x)=0 which implies x4+27=1 This leads to x4=−26 which has no real solutions. There are no xintercepts.

4. Analyze the first derivative to find extrema. Using the chain rule:

d(ƒ(x))/d(x)=(4*x3)/(x4+27)

Setting the derivative to zero gives 4*x3=0 so x=0 For x<0 ƒ(x)′<0 (decreasing). For x>0 ƒ(x)′>0 (increasing). There is a local minimum at (0,ln(27))

5. Analyze the second derivative for concavity and inflection points. Using the quotient rule:

d2(ƒ(x))/(d(x)2)=(12*x2*(x4+27)−4*x3*(4*x3))/((x4+27)2)

d2(ƒ(x))/(d(x)2)=(12*x6+324*x2−16*x6)/((x4+27)2)

d2(ƒ(x))/(d(x)2)=(324*x2−4*x6)/((x4+27)2)

Setting the numerator to zero: 4*x2*(81−x4)=0 This gives x=0 and x=±3 Testing intervals shows the graph is concave up on (−3,3) and concave down on (−∞,−3) and (3,∞) Inflection points occur at x=±3

6. Examine end behavior. As x→∞ or x→−∞ x4+27→∞ so ƒ(x)→∞ There are no horizontal asymptotes.

Final Answer

ƒ(x)=ln(x4+27)

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