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aniked [119]
3 years ago
9

Verify that f and g are inverses functions using composition , show your steps

Mathematics
1 answer:
pshichka [43]3 years ago
5 0

Answer:

See explanations below

Step-by-step explanation:

Given the functions

f(x) = 12x - 12

g(x) = x/12 - 1

To show they are inverses, we, must show that f(g(x)) = g(f(x))

f(g(x)) = f(x/12 - 1)

Replace x with x/12 - 1 into f(x)

f(g(x)) =12((x-12)/12) - 11

f(g(x)) =  x-1 - 1

f(g(x)) =x - 2

Similarly for g(f(x))

g(f(x)) = g(12x-12)

g(f(x)) =(12x-12)/12 - 1

12(x-1)/12 - 1

x-1 - 1

x - 2

Since f(g(x)) = g(f(x)) = x -2, hence they are inverses of each other

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a) The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }.

b) The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}.

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}.

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable (y, t) in each side of the identity:

6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t (1)

6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C

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-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C

If we know that x = 0 and y = 1<em>, </em>then the integration constant is C = -2.

The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }. \blacksquare

b) In this case we need to solve a first order ordinary differential equation of the following form:

\frac{dy}{dx} + p(x) \cdot y = q(x) (2)

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The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}. \blacksquare

The solution for (2) is presented below:

y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C (4)

Where C is the integration constant.

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