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Katyanochek1 [597]
3 years ago
15

If f(x) = 1/3x+13, then f^-1(x) =??​

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Step-by-step explanation:

Given

f(x) = 1/3x + 13

f^-1(x) = ?

Now

Let y = f(x)

or y = 1/3x + 13

Interchanging the role of x and y

x = 1/3y + 13

x - 13 = 1/3y

y = <u>x </u><u>-</u><u> </u><u>1</u><u>3</u>

3

Therefore f^-1(x) = <u>x </u><u>-</u><u> </u><u>1</u><u>3</u><u> </u>

3

Hope it will help you.

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Soloha48 [4]

Answer:

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2 years ago
Find the function y = f(t) passing through the point (0, 18) with the given first derivative.
monitta

Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
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Equality Properties

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<u>Algebra I</u>

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<u>Calculus</u>

Derivatives

Derivative Notation

Antiderivatives - Integrals

Integration Constant C

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Integration Property [Multiplied Constant]:                                                             \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

\displaystyle \frac{dy}{dt} = \frac{1}{8} t

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

  1. [Derivative] Rewrite:                                                                                         \displaystyle dy = \frac{1}{8} t\ dt
  2. [Equality Property] Integrate both sides:                                                        \displaystyle \int dy = \int {\frac{1}{8} t} \, dt
  3. [Left Integral] Integrate [Integration Rule - Reverse Power Rule]:                 \displaystyle y = \int {\frac{1}{8} t} \, dt
  4. [Right Integral] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle y = \frac{1}{8}\int {t} \, dt
  5. [Right Integral] Integrate [Integration Rule - Reverse Power Rule]:              \displaystyle y = \frac{1}{8}(\frac{t^2}{2}) + C
  6. Multiply:                                                                                                             \displaystyle y = \frac{t^2}{16} + C

<u>Step 3: Find Particular Solution</u>

  1. Substitute in point [Function]:                                                                         \displaystyle 18 = \frac{0^2}{16} + C
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  3. Add:                                                                                                                   \displaystyle 18 = C
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  5. Substitute in <em>C</em> [Function]:                                                                                \displaystyle y = \frac{t^2}{16} + 18

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

Unit: Integration

Book: College Calculus 10e

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