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otez555 [7]
3 years ago
10

For all the values of x, f(x) = 2x(squared) g(x) = x+1 solve fg(x)=gf(x)

Mathematics
1 answer:
bulgar [2K]3 years ago
3 0

Answer:

Value of x for fg(x)=gf(x) at <em>f(x) = 2x²</em> and <em>g(x) = x+1</em> is x = -1/4 .

Step-by-step explanation:

We have, f(x) = 2x² and g(x) = x+1

Now, f(g(x)) = g(f(x))

⇒ <em>2(x+1)² = 2x² + 1</em>

<em>⇒ 2( x² + 2x + 1 ) = 2x² + 1</em>

<em>⇒ 2x² + 4x + 2 = 2x² + 1</em>

<em>⇒ 4x = -1</em>

<em>⇒ </em><em>x = -1/4</em>

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Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] &lt; y0 &lt; [infinity](a) Sketch the graph
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Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of  f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0  and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane  is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of  f(y) versus y

Looking at the given differential equation

       \frac{dy}{dt} = e^{y} - 1 for -∞ < y_{o} < ∞

 We can say let \frac{dy}{dt} = f(y) =e^{y} - 1

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

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Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where  f(y) = 0.

In each of the intervals bounded  by the equilibrium draw an upward

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      This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For  below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper  

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Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium  that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)

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