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

Answer:

Concave Up Interval: $(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)$

Concave Down Interval: $(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )$

General Formulas and Concepts:

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

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

Step-by-step explanation:

Step 1: Define

$f(x)=\frac{3}{1+x^2}$

Step 2: Find 2nd Derivative

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

Step 3: Find P.P.I

• Set f"(x) equal to zero:                    $0=\frac{6(3x^2-1)}{(1+x^2)^3}$

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:                                         $\frac{1}{3} =x^2$
5. Square root:                                   $\pm \sqrt{\frac{1}{3}} =x$
6. Simplify:                                          $\pm \frac{\sqrt{3}}{3} =x$
7. Rewrite:                                          $x= \pm \frac{\sqrt{3}}{3}$

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 $x= \pm \frac{\sqrt{3}}{3}$ (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:                    $f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}$
2. Exponents:                   $f"(x)=\frac{6(3(1)-1)}{(1+1)^3}$
3. Multiply:                        $f"(x)=\frac{6(3-1)}{(1+1)^3}$
4. Subtract/Add:              $f"(x)=\frac{6(2)}{(2)^3}$
5. Exponents:                  $f"(x)=\frac{6(2)}{8}$
6. Multiply:                       $f"(x)=\frac{12}{8}$
7. Simplify:                       $f"(x)=\frac{3}{2}$

This means that the graph f(x) is concave up before $x=\frac{-\sqrt{3}}{3}$.

x = 0

1. Substitute:                    $f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}$
2. Exponents:                   $f"(x)=\frac{6(3(0)-1)}{(1+0)^3}$
3. Multiply:                       $f"(x)=\frac{6(0-1)}{(1+0)^3}$
4. Subtract/Add:              $f"(x)=\frac{6(-1)}{(1)^3}$
5. Exponents:                  $f"(x)=\frac{6(-1)}{1}$
6. Multiply:                       $f"(x)=\frac{-6}{1}$
7. Divide:                         $f"(x)=-6$

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

x = 1

1. Substitute:                    $f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}$
2. Exponents:                   $f"(x)=\frac{6(3(1)-1)}{(1+1)^3}$
3. Multiply:                       $f"(x)=\frac{6(3-1)}{(1+1)^3}$
4. Subtract/Add:              $f"(x)=\frac{6(2)}{(2)^3}$
5. Exponents:                  $f"(x)=\frac{6(2)}{8}$
6. Multiply:                       $f"(x)=\frac{12}{8}$
7. Simplify:                       $f"(x)=\frac{3}{2}$

This means that the graph f(x) is concave up after $x=\frac{\sqrt{3}}{3}$.

Step 5: Identify

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

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

$x=\frac{-\sqrt{3}}{3}$

1. Substitute in P.I into f(x):                    $f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}$
2. Evaluate Exponents:                          $f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }$
3. Add:                                                    $f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }$
4. Divide:                                                $f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}$

$x=\frac{\sqrt{3}}{3}$

1. Substitute in P.I into f(x):                    $f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}$
2. Evaluate Exponents:                          $f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }$
3. Add:                                                    $f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }$
4. Divide:                                                $f(\frac{\sqrt{3}}{3} )=\frac{9}{4}$

Step 6: Define Intervals

We know that before f(x) reaches $x=\frac{-\sqrt{3}}{3}$, the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that after f(x) passes $x=\frac{\sqrt{3}}{3}$, the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: $(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)$

We know that after f(x) passes $x=\frac{-\sqrt{3}}{3}$ , the graph is concave up until $x=\frac{\sqrt{3}}{3}$. We used the 2nd Derivative Test to confirm this.

Concave Down Interval: $(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )$