Question

Can anybody help plzz?? 65 points

Answer 1

Answer:

$\frac{dy}{dx} =\frac{-8}{x^2} +2$

$\frac{d^2y}{dx^2} =\frac{16}{x^3}$

Stationary Points: See below.

General Formulas and Concepts:

Pre-Algebra

• Equality Properties

Calculus

Derivative Notation dy/dx

Derivative of a Constant equals 0.

Stationary Points are where the derivative is equal to 0.

• 1st Derivative Test - Tells us if the function f(x) has relative max or mins. Critical Numbers occur when f'(x) = 0 or f'(x) = undef
• 2nd Derivative Test - Tells us the function f(x)'s concavity behavior. Possible Points of Inflection/Points of Inflection occur when f"(x) = 0 or f"(x) = undef

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

Step 1: Define

$f(x)=\frac{8}{x} +2x$

Step 2: Find 1st Derivative (dy/dx)

1. Quotient Rule [Basic Power]:                    $f'(x)=\frac{0(x)-1(8)}{x^2} +2x$
2. Simplify:                                                      $f'(x)=\frac{-8}{x^2} +2x$
3. Basic Power Rule:                                     $f'(x)=\frac{-8}{x^2} +1 \cdot 2x^{1-1}$
4. Simplify:                                                     $f'(x)=\frac{-8}{x^2} +2$

Step 3: 1st Derivative Test

1. Set 1st Derivative equal to 0:                    $0=\frac{-8}{x^2} +2$
2. Subtract 2 on both sides:                         $-2=\frac{-8}{x^2}$
3. Multiply x² on both sides:                         $-2x^2=-8$
4. Divide -2 on both sides:                           $x^2=4$
5. Square root both sides:                            $x= \pm 2$

Our Critical Points (stationary points for rel max/min) are -2 and 2.

Step 4: Find 2nd Derivative (d²y/dx²)

1. Define:                                                      $f'(x)=\frac{-8}{x^2} +2$
2. Quotient Rule [Basic Power]:                  $f''(x)=\frac{0(x^2)-2x(-8)}{(x^2)^2} +2$
3. Simplify:                                                    $f''(x)=\frac{16}{x^3} +2$
4. Basic Power Rule:                                    $f''(x)=\frac{16}{x^3}$

Step 5: 2nd Derivative Test

1. Set 2nd Derivative equal to 0:                    $0=\frac{16}{x^3}$
2. Solve for x:                                                    $x = 0$

Our Possible Point of Inflection (stationary points for concavity) is 0.

Step 6: Find coordinates

Plug in the C.N and P.P.I into f(x) to find coordinate points.

x = -2

1. Substitute:                    $f(-2)=\frac{8}{-2} +2(-2)$
2. Divide/Multiply:            $f(-2)=-4-4$
3. Subtract:                       $f(-2)=-8$

x = 2

1. Substitute:                    $f(2)=\frac{8}{2} +2(2)$
2. Divide/Multiply:            $f(2)=4 +4$
3. Add:                              $f(2)=8$

x = 0

1. Substitute:                    $f(0)=\frac{8}{0} +2(0)$
2. Evaluate:                      $f(0)=\text{unde} \text{fined}$

Step 7: Identify Behavior

See Attachment.

Point (-2, -8) is a relative max because f'(x) changes signs from + to -.

Point (2, 8) is a relative min because f'(x) changes signs from - to +.

When x = 0, there is a concavity change because f"(x) changes signs from - to +.