Graph f(x) = natural log of x^2+25

Problem

ƒ(x)=ln(x2+25)

Solution

1. Identify the domain of the function. Since x2+25 is always greater than or equal to 25 for all real x the argument of the natural log is always positive. Thus, the domain is (−∞,∞)

2. Determine the symmetry of the function. Substitute −x for x ƒ*(−x)=ln((−x)2+25)=ln(x2+25) Since ƒ(x)=ƒ*(−x) the function is even and symmetric about the yaxis.

3. Find the intercepts. For the yintercept, set x=0 ƒ(0)=ln(0+25)=ln(25)≈3.22 For xintercepts, set ƒ(x)=0 ln(x2+25)=0⇒x2+25=1⇒x2=−24 which has no real solutions. There are no xintercepts.

4. Find the derivative to determine extrema and intervals of increase/decrease. Use the chain rule:

d(ln(x2+25))/d(x)=(2*x)/(x2+25)

5. Locate critical points by setting ƒ(x)′=0 This occurs at x=0 For x<0 ƒ(x)′<0 (decreasing). For x>0 ƒ(x)′>0 (increasing). There is a local minimum at (0,ln(25))

6. Find the second derivative to determine concavity and inflection points. Use the quotient rule:

d()/d(x)(2*x)/(x2+25)=(2*(x2+25)−2*x*(2*x))/((x2+25)2)

ƒ(x)″=(50−2*x2)/((x2+25)2)

7. Identify inflection points by setting ƒ(x)″=0 This occurs when 50−2*x2=0 so x2=25 which means x=5 and x=−5 The graph is concave up on (−5,5) and concave down on (−∞,−5)∪(5,∞)

8. Analyze end behavior. As x→∞ or x→−∞ x2+25→∞ so ƒ(x)→∞ The graph opens upward indefinitely.

Final Answer

ƒ(x)=ln(x2+25)

Want more problems? Check here!