Mathematics, 25.11.2020 23:10 leannamat2106

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 f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to f

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Answer from: ayoismeisalex

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) = cxⁿf’(x) = c·nxⁿ⁻¹

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

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 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

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

Step 2: Find 2nd Derivative

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

Solve Numerator: 0=6(3x^2-1)

Divide 6:

Add 1:

Divide 3: $\frac{1}{3} =x^2$ Square root: $\pm \sqrt{\frac{1}{3}} =x$ Simplify: $\pm \frac{\sqrt{3}}{3} =x$ Rewrite: $x= \pm \frac{\sqrt{3}}{3}$

Case 2: f" is undefined

Solve Denominator: 0=(1+x^2)^3

Cube root: 0=1+x^2

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

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

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

x = 0

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

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

x = 1

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

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

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

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

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

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

Answer from: Quest

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it

Me out and tell me the correct answer if you dont know dont comment

Answer from: Quest

486in is correct

step-by-step explanation:

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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...