Question

The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to find the intervals on which f is concave upward or concave downward and to find the inflection points of f.

Answer 1

Concave Up Interval:
`(- \infty,(-√(3) )/(3) )U((√(3) )/(3) , \infty)`

Concave Down Interval:
`((-√(3) )/(3), (√(3) )/(3) )`

General Formulas and Concepts:

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Quotient Rule:
`(d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))`

Chain Rule:
`(d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Second Derivative Test:

• Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefined
• Points of Inflection (P.I) - Actual x-values when the graph f(x) changes concavity
• Number Line Test - Helps us determine whether a P.P.I is a P.I

Explanation:

Step 1: Define

`f(x)=(3)/(1+x^2)`

Step 2: Find 2nd Derivative

1. 1st Derivative [Quotient/Chain/Basic]:
   `f'(x)=(0(1+x^2)-2x \cdot 3)/((1+x^2)^2)`
2. Simplify 1st Derivative:
   `f'(x)=(-6x)/((1+x^2)^2)`
3. 2nd Derivative [Quotient/Chain/Basic]:
   `f`
4. Simplify 2nd Derivative:
   `f`

Step 3: Find P.P.I

• Set f"(x) equal to zero:
  `0=(6(3x^2-1))/((1+x^2)^3)`

Case 1: f" is 0

1. Solve Numerator:
   `0=6(3x^2-1)`
2. Divide 6:
   `0=3x^2-1`
3. Add 1:
   `1=3x^2`
4. Divide 3:
   `(1)/(3) =x^2`
5. Square root:
   `\pm \sqrt{(1)/(3)} =x`
6. Simplify:
   `\pm (√(3))/(3) =x`
7. Rewrite:
   `x= \pm (√(3))/(3)`

Case 2: f" is undefined

1. Solve Denominator:
   `0=(1+x^2)^3`
2. Cube root:
   `0=1+x^2`
3. Subtract 1:
   `-1=x^2`

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is
`x= \pm (√(3))/(3)` (x ≈ ±0.57735).

Step 4: Number Line Test

See Attachment.

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = -1

1. Substitute:
   `f`
2. Exponents:
   `f`
3. Multiply:
   `f`
4. Subtract/Add:
   `f`
5. Exponents:
   `f`
6. Multiply:
   `f`
7. Simplify:
   `f`

This means that the graph f(x) is concave up before
`x=(-√(3))/(3)`.

x = 0

1. Substitute:
   `f`
2. Exponents:
   `f`
3. Multiply:
   `f`
4. Subtract/Add:
   `f`
5. Exponents:
   `f`
6. Multiply:
   `f`
7. Divide:
   `f`

This means that the graph f(x) is concave down between and .

x = 1

1. Substitute:
   `f`
2. Exponents:
   `f`
3. Multiply:
   `f`
4. Subtract/Add:
   `f`
5. Exponents:
   `f`
6. Multiply:
   `f`
7. Simplify:
   `f`

This means that the graph f(x) is concave up after
`x=(√(3))/(3)`.

Step 5: Identify

Since f"(x) changes concavity from positive to negative at
`x=(-√(3))/(3)` and changes from negative to positive at
`x=(√(3))/(3)`, then we know that the P.P.I's
`x= \pm (√(3))/(3)` are actually P.I's.

Let's find what actual point on f(x) when the concavity changes.

`x=(-√(3))/(3)`

1. Substitute in P.I into f(x):
   `f((-√(3))/(3) )=(3)/(1+((-√(3) )/(3) )^2)`
2. Evaluate Exponents:
   `f((-√(3))/(3) )=(3)/(1+(1)/(3) )`
3. Add:
   `f((-√(3))/(3) )=(3)/((4)/(3) )`
4. Divide:
   `f((-√(3))/(3) )=(9)/(4)`

`x=(√(3))/(3)`

1. Substitute in P.I into f(x):
   `f((√(3))/(3) )=(3)/(1+((√(3) )/(3) )^2)`
2. Evaluate Exponents:
   `f((√(3))/(3) )=(3)/(1+(1)/(3) )`
3. Add:
   `f((√(3))/(3) )=(3)/((4)/(3) )`
4. Divide:
   `f((√(3))/(3) )=(9)/(4)`

Step 6: Define Intervals

We know that before f(x) reaches
`x=(-√(3))/(3)`, the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that after f(x) passes
`x=(√(3))/(3)`, the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval:
`(- \infty,(-√(3) )/(3) )U((√(3) )/(3) , \infty)`

We know that after f(x) passes
`x=(-√(3))/(3)` , the graph is concave up until
`x=(√(3))/(3)`. We used the 2nd Derivative Test to confirm this.

Concave Down Interval:
`((-√(3) )/(3), (√(3) )/(3) )`