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3 years ago

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If f(x) = 4x2 - 3 and g(x) = x+1, find (f - g)(x).

1 answer:

tresset_1 [31]3 years ago

4 0

Answer:

C. 4x² - x - 4

Step-by-step explanation:

Step 1: Define functions

f(x) = 4x² - 3

g(x) = x + 1

Step 2: Subtract

(f - g)(x) = 4x² - 3 - (x + 1)

(f - g)(x) = 4x² - 3 - x - 1

(f - g)(x) = 4x² - x - 4

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Can anybody help plzz?? 65 points

Yakvenalex [24]

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:

<u>Pre-Algebra</u>

Equality Properties

<u>Calculus</u>

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:

<u>Step 1: Define</u>

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

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

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

<u>Step 3: 1st Derivative Test</u>

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

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

<u>Step 4: Find 2nd Derivative (d²y/dx²)</u>

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

<u>Step 5: 2nd Derivative Test</u>

Set 2nd Derivative equal to 0: $0=\frac{16}{x^3}$
Solve for <em>x</em>: $x = 0$

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

<u>Step 6: Find coordinates</u>

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

x = -2

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

x = 2

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

x = 0

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

<u>Step 7: Identify Behavior</u>

<em>See Attachment.</em>

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 +.

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