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mafiozo [28]
3 years ago
15

A commercial comes on for an arthritis cream. Lachlan has been having some pain in his knees lately, so he's both motivated and

able to attend to the commercial. Therefore, he is more likely to use the ________ route to persuasion.A. peripheralB. heuristicC. centralD. relevant
Health
1 answer:
vova2212 [387]3 years ago
4 0

Answer: C. central

Explanation:

The central route of persuasion emphasizes over the fact that content which is directly related to the fact and relevant is shown to the person. Then the person is more likely to persuade to the change in his or her behavior. The fact may attract the attention of the audience, may loose credibility and have a definitive argument.

Here, in the given situation, the commercial is directly related to Lachlan. As this person is experiencing pain in the knees which can be probable cause of arthritis. Thus the commercial is persuading the Lachlan centrally.

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Differentiate the following functions (i) x(1+x)^3​
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Answer:

\displaystyle y' = (1 + x)^2(4x + 1)

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
  • Functions
  • Function Notation
  • Factoring

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]:                                                                \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                                \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

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

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

y = x(1 + x)³

<u>Step 2: Differentiate</u>

  1. Product Rule [Derivative Rule - Chain Rule]:                                                  \displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]
  2. Derivative Property [Addition/Subtraction]:                                                    \displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])
  3. Basic Power Rule:                                                                                             \displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})
  4. Simplify:                                                                                                             \displaystyle y' = (1 + x)^3 + 3x(1 + x)^2
  5. Factor:                                                                                                               \displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]
  6. Combine like terms:                                                                                         \displaystyle y' = (1 + x)^2(4x + 1)

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

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