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zaharov [31]
3 years ago
11

The function f(x) = −16x2 + 112 models the height, in feet, of an

Mathematics
1 answer:
natima [27]3 years ago
3 0

For this, just plug in 1.5 into the x variable of this function and solve as such:

f(1.5)=-16*1.5^2+112\\f(1.5)=-16*2.25+112\\f(1.5)=-36+112\\f(1.5)=76

<u>The height of the object was at 76 ft.</u>

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If f(x) = 9x10 tan−1x, find f '(x).
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Answer:

\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}

General Formulas and Concepts:

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Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)  

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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = 9x^{10} \tan^{-1}(x)

<u>Step 2: Differentiate</u>

  1. [Function] Derivative Rule [Product Rule]:                                                   \displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
  2. Rewrite [Derivative Property - Multiplied Constant]:                                  \displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
  3. Basic Power Rule:                                                                                         \displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
  4. Arctrig Derivative:                                                                                         \displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}

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

Unit: Differentiation

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