contestada

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.

The graph of fx 31x2 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 a class=

Respuesta :

Space

Answer:

Concave Up Interval: [tex](- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)[/tex]

Concave Down Interval: [tex](\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )[/tex]

General Formulas and Concepts:

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

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

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

[tex]f(x)=\frac{3}{1+x^2}[/tex]

Step 2: Find 2nd Derivative

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

Step 3: Find P.P.I

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

Case 1: f" is 0

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

Case 2: f" is undefined

  1. Solve Denominator:                    [tex]0=(1+x^2)^3[/tex]
  2. Cube root:                                   [tex]0=1+x^2[/tex]
  3. Subtract 1:                                    [tex]-1=x^2[/tex]

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is [tex]x= \pm \frac{\sqrt{3}}{3}[/tex] (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:                    [tex]f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}[/tex]
  2. Exponents:                   [tex]f"(x)=\frac{6(3(1)-1)}{(1+1)^3}[/tex]
  3. Multiply:                        [tex]f"(x)=\frac{6(3-1)}{(1+1)^3}[/tex]
  4. Subtract/Add:              [tex]f"(x)=\frac{6(2)}{(2)^3}[/tex]
  5. Exponents:                  [tex]f"(x)=\frac{6(2)}{8}[/tex]
  6. Multiply:                       [tex]f"(x)=\frac{12}{8}[/tex]
  7. Simplify:                       [tex]f"(x)=\frac{3}{2}[/tex]

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

x = 0

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

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

x = 1

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

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

Step 5: Identify

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

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

[tex]x=\frac{-\sqrt{3}}{3}[/tex]

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

[tex]x=\frac{\sqrt{3}}{3}[/tex]

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

Step 6: Define Intervals

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

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

Concave Up Interval: [tex](- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)[/tex]

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

Concave Down Interval: [tex](\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )[/tex]

Ver imagen Space

Otras preguntas