Sketch the region enclosed by the given curves.

y = 4/x, y = 16x, y =1/4x, x > 0

Find the area.

To sketch the region enclosed by the given curves, we need to plot the curves on a coordinate plane and identify the boundaries of the enclosed region.

1. Start by plotting the curve y = 4/x. This curve is a hyperbola. As x approaches infinity, y approaches 0. As x approaches 0, y approaches infinity.

2. Next, plot the curve y = 16x. This curve is a straight line passing through the origin (0,0) with a slope of 16. As x increases, y also increases.

3. Plot the curve y = 1/4x. This curve is also a hyperbola. As x approaches infinity, y approaches 0. As x approaches 0, y approaches infinity.

4. Now, consider the boundaries of the enclosed region. The curves intersect at certain points, which define the boundaries of the region. To find these points, we can set the equations for the curves equal to each other and solve for x and y.

Setting y = 4/x and y = 16x: 4/x = 16x

Rearranging the equation: 4 = 16x^2

Solving for x: x^2 = 4/16 => x^2 = 1/4 => x = ±1/2

Plugging in the obtained x values into either of the equations, we can find the corresponding y values.

For x = 1/2: y = 4/(1/2) => y = 8
For x = -1/2: y = 4/(-1/2) => y = -8

So the points of intersection are (1/2, 8) and (-1/2, -8).

5. Finally, we shade the enclosed region between the curves. The region is bound by the x-axis, the curves y = 4/x and y = 16x, and the lines x = 0 and x = 1/2. The region will look like a triangle with a curved side.

To find the area of this region, you can split it into two components: the triangle and the curved side.

Area of the triangle:
Base = 1/2
Height = 8
Area = (Base * Height) / 2 = (1/2 * 8) / 2 = 4

Area of the curved side:
The curved side can be split into two parts, one from y = 4/x and one from y = 1/4x. To find the area, we need to integrate the difference between the functions from x = -1/2 to x = 1/2.

∫[(4/x) - (1/4x)] dx

Integrating the above expression will give you the area of the curved side.

Please note that the above process provides a general approach, and the detailed calculations may vary depending on the level of accuracy and specific requirements.

To sketch the region enclosed by the given curves, we need to first plot each curve and identify their points of intersection.

1. Start by plotting the curve y = 4/x. Since x > 0, we can consider the positive values of x. As x approaches 0, y tends to infinity. As x increases, y decreases. Plot points for different positive values of x to get an idea of how the curve looks.

2. Next, plot the curve y = 16x. This is a straight line that passes through the origin with a positive slope.

3. Finally, plot the curve y = 1/4x. This is another straight line with a positive slope passing through the origin.

Now, identify the points of intersection of these curves. To do this, set the equations equal to each other and solve for the values of x and y.

For example, set y = 4/x equal to y = 16x:

4/x = 16x
4 = 16x^2
x^2 = 1/4
x = 1/2 or x = -1/2 (but we discard the negative value since x > 0)
When x = 1/2, y = 4/(1/2) = 8.

Similarly, set y = 4/x equal to y = 1/4x:

4/x = 1/4x
4 = x^2
x = 2 or x = -2 (again, we discard the negative value)
When x = 2, y = 4/(2) = 2.

So the three curves intersect at the points (1/2, 8) and (2, 2).

Now, we can sketch the region enclosed by the curves, which is bounded by the x-axis, and the two lines (16x and 1/4x) along with y = 4/x between the points of intersection.

To find the area of this region, we need to integrate the y-values over the range of x-values.

The integral of y = 4/x from x = 1/2 to x = 2 gives us the area enclosed by the curves.

So, to find the area, integrate the function y = 4/x with respect to x from x = 1/2 to x = 2.

The area can be calculated as:

Area = ∫[1/2, 2] (4/x) dx

Evaluating this integral will give us the required area within the region enclosed by the given curves.

y = 4/x and y = 16x intersect at (1/2 , 8)

and y = 4/x and y = x/4 intersect at (4,1)

make a sketch and you will see that you have to do it two parts
∫(16x -x/4)dx from 0 to 1/2 + ∫(4/x - x/4)dx from 1/2 to 4

recall that ∫ 1/x dx = ln x

the rest is mostly arithmetic