Suppose that f(x)=ln(3+x^2). Use interval notation to state when f(x) is concave up and concave down. Then find all inflection points for f(x).
f' = 2x/(3+x^2)
f" = 2(3-x^2)/(3+x^2)^2
inflection where f" = 0. That is, where
x^2 = 3
concave up where f" > 0
To determine when a function is concave up or concave down, we need to analyze its second derivative.
Step 1: Find the first derivative of f(x):
To find the first derivative of f(x), we can use the chain rule. The derivative of ln(u) with respect to x is 1/u multiplied by the derivative of u with respect to x. In this case, u = 3 + x^2.
f'(x) = (1 / (3 + x^2)) * (2x)
= 2x / (3 + x^2)
Step 2: Find the second derivative of f(x):
To find the second derivative, we differentiate f'(x) with respect to x.
Using the quotient rule, we obtain:
f''(x) = [(2 * (3 + x^2)) - (2x * 2x)] / (3 + x^2)^2
= (6 + 2x^2 - 4x^2) / (3 + x^2)^2
= (6 - 2x^2) / (3 + x^2)^2
Step 3: Determine when f(x) is concave up or concave down:
For f(x) to be concave up, the second derivative f''(x) must be positive.
For f(x) to be concave down, the second derivative f''(x) must be negative.
To find the intervals where f(x) is concave up and concave down, we need to solve the inequality f''(x) > 0 for concave up, and f''(x) < 0 for concave down.
Let's determine the sign of f''(x):
f''(x) = (6 - 2x^2) / (3 + x^2)^2
For f''(x) to be positive:
6 - 2x^2 > 0
- 2x^2 > -6
x^2 < 3
Taking the square root of both sides, we get:
-√3 < x < √3
For f''(x) to be negative, it follows that x must be outside the interval -√3 < x < √3.
Hence, the function f(x) is concave up on the interval (-√3, √3), and concave down when x < -√3 or x > √3.
Step 4: Find the inflection points:
An inflection point occurs when the concavity of the function changes. In other words, it is a point where the second derivative changes sign.
To find the inflection points, we set f''(x) = 0 and solve for x:
(6 - 2x^2) / (3 + x^2)^2 = 0
Cross-multiplying, we have:
6 - 2x^2 = 0
6 = 2x^2
x^2 = 3
x = ±√3
Therefore, the inflection points for f(x) are x = -√3 and x = √3.
In summary,
- f(x) is concave up on the interval (-√3, √3).
- f(x) is concave down when x < -√3 or x > √3.
- Inflection points: x = -√3 and x = √3.