Concave Up Interval: ![(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)](/tpl/images/0929/2566/a5c60.png)
Concave Down Interval: ![(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )](/tpl/images/0929/2566/4c727.png)
General Formulas and Concepts:
Calculus
Derivative of a Constant is 0.
Basic Power Rule:
f(x) = cxⁿf’(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)}](/tpl/images/0929/2566/a2a0e.png)
Chain Rule: ![\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](/tpl/images/0929/2566/44454.png)
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 undefinedPoints of Inflection (P.I) - Actual x-values when the graph f(x) changes concavityNumber 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}](/tpl/images/0929/2566/162bb.png)
Step 2: Find 2nd Derivative
1st Derivative [Quotient/Chain/Basic]:
![f'(x)=\frac{0(1+x^2)-2x \cdot 3}{(1+x^2)^2}](/tpl/images/0929/2566/01a8c.png)
Simplify 1st Derivative:
![f'(x)=\frac{-6x}{(1+x^2)^2}](/tpl/images/0929/2566/91817.png)
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}](/tpl/images/0929/2566/88a93.png)
Simplify 2nd Derivative:
![f"(x)=\frac{6(3x^2-1)}{(1+x^2)^3}](/tpl/images/0929/2566/7d67d.png)
Step 3: Find P.P.I
Set f"(x) equal to zero:
![0=\frac{6(3x^2-1)}{(1+x^2)^3}](/tpl/images/0929/2566/0cdde.png)
Case 1: f" is 0
Solve Numerator:
![0=6(3x^2-1)](/tpl/images/0929/2566/58363.png)
Divide 6:
![0=3x^2-1](/tpl/images/0929/2566/04368.png)
Add 1:
![1=3x^2](/tpl/images/0929/2566/8e8ff.png)
Divide 3:
![\frac{1}{3} =x^2](/tpl/images/0929/2566/cf55b.png)
Square root:
![\pm \sqrt{\frac{1}{3}} =x](/tpl/images/0929/2566/bcf7e.png)
Simplify:
![\pm \frac{\sqrt{3}}{3} =x](/tpl/images/0929/2566/73ec8.png)
Rewrite:
![x= \pm \frac{\sqrt{3}}{3}](/tpl/images/0929/2566/42110.png)
Case 2: f" is undefined
Solve Denominator:
![0=(1+x^2)^3](/tpl/images/0929/2566/c54f0.png)
Cube root:
![0=1+x^2](/tpl/images/0929/2566/fb227.png)
Subtract 1:
![-1=x^2](/tpl/images/0929/2566/67bef.png)
We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is
(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
Substitute:
![f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}](/tpl/images/0929/2566/98212.png)
Exponents:
![f"(x)=\frac{6(3(1)-1)}{(1+1)^3}](/tpl/images/0929/2566/26413.png)
Multiply:
![f"(x)=\frac{6(3-1)}{(1+1)^3}](/tpl/images/0929/2566/a5f80.png)
Subtract/Add:
![f"(x)=\frac{6(2)}{(2)^3}](/tpl/images/0929/2566/bcf24.png)
Exponents:
![f"(x)=\frac{6(2)}{8}](/tpl/images/0929/2566/0faa5.png)
Multiply:
![f"(x)=\frac{12}{8}](/tpl/images/0929/2566/fa652.png)
Simplify:
![f"(x)=\frac{3}{2}](/tpl/images/0929/2566/c71c7.png)
This means that the graph f(x) is concave up before
.
x = 0
Substitute:
![f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}](/tpl/images/0929/2566/22770.png)
Exponents:
![f"(x)=\frac{6(3(0)-1)}{(1+0)^3}](/tpl/images/0929/2566/5ef01.png)
Multiply:
![f"(x)=\frac{6(0-1)}{(1+0)^3}](/tpl/images/0929/2566/cbba3.png)
Subtract/Add:
![f"(x)=\frac{6(-1)}{(1)^3}](/tpl/images/0929/2566/bab88.png)
Exponents:
![f"(x)=\frac{6(-1)}{1}](/tpl/images/0929/2566/e3a4b.png)
Multiply:
![f"(x)=\frac{-6}{1}](/tpl/images/0929/2566/4b24e.png)
Divide:
![f"(x)=-6](/tpl/images/0929/2566/b0813.png)
This means that the graph f(x) is concave down between and .
x = 1
Substitute:
![f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}](/tpl/images/0929/2566/61968.png)
Exponents:
![f"(x)=\frac{6(3(1)-1)}{(1+1)^3}](/tpl/images/0929/2566/26413.png)
Multiply:
![f"(x)=\frac{6(3-1)}{(1+1)^3}](/tpl/images/0929/2566/a5f80.png)
Subtract/Add:
![f"(x)=\frac{6(2)}{(2)^3}](/tpl/images/0929/2566/bcf24.png)
Exponents:
![f"(x)=\frac{6(2)}{8}](/tpl/images/0929/2566/0faa5.png)
Multiply:
![f"(x)=\frac{12}{8}](/tpl/images/0929/2566/fa652.png)
Simplify:
![f"(x)=\frac{3}{2}](/tpl/images/0929/2566/c71c7.png)
This means that the graph f(x) is concave up after
.
Step 5: Identify
Since f"(x) changes concavity from positive to negative at
and changes from negative to positive at
, then we know that the P.P.I's
are actually P.I's.
Let's find what actual point on f(x) when the concavity changes.
![x=\frac{-\sqrt{3}}{3}](/tpl/images/0929/2566/0fc0d.png)
Substitute in P.I into f(x):
![f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}](/tpl/images/0929/2566/d888d.png)
Evaluate Exponents:
![f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }](/tpl/images/0929/2566/5a75d.png)
Add:
![f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }](/tpl/images/0929/2566/fc5cf.png)
Divide:
![f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}](/tpl/images/0929/2566/bb327.png)
![x=\frac{\sqrt{3}}{3}](/tpl/images/0929/2566/afdef.png)
Substitute in P.I into f(x):
![f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}](/tpl/images/0929/2566/e3c04.png)
Evaluate Exponents:
![f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }](/tpl/images/0929/2566/bbebb.png)
Add:
![f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }](/tpl/images/0929/2566/ef3c5.png)
Divide:
![f(\frac{\sqrt{3}}{3} )=\frac{9}{4}](/tpl/images/0929/2566/444aa.png)
Step 6: Define Intervals
We know that before f(x) reaches
, the graph is concave up. We used the 2nd Derivative Test to confirm this.
We know that after f(x) passes
, 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)](/tpl/images/0929/2566/a5c60.png)
We know that after f(x) passes
, the graph is concave up until
. We used the 2nd Derivative Test to confirm this.
Concave Down Interval: ![(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )](/tpl/images/0929/2566/4c727.png)
![The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to fi](/tpl/images/0929/2566/03fb9.jpg)