1. Conslaer the Curve y =JX) = 2"-1.
A. Find the exact area of the region in the first quadrant bounded by the curves y = fx) = 2-1
and y = x. ("Exact area" means no calculator numbers.)
B. Find the inverse function y =flx).
C. Using part A and the notion of symmetry between a function and its inverse, find the exact
area of the region in the first quadrant bounded by the curves y = flx) and y = x. Explain your
reasoning. (Hint: Think "graphically" and little or no math will need to be done!)
No idea what y =JX) = 2"-1 is.
If you mean y = 2^x - 1
then
(A) the area on the interval [1,2] is
∫[1,2] x-(2^x - 1) dx
(B) f-1(x) = log2(x+1)
(C) see what you can do with this.
If you actually meant y = 2^(x-1) then make the appropriate adjustments
To find the exact area of the region in the first quadrant bounded by the curves y = f(x) = 2^(-1) and y = x, we can follow the steps below:
A. Find the exact area between the curves y = f(x) = 2^(-1) and y = x:
1. Set the two equations equal to each other and solve for x:
2^(-1) = x
Since 2^(-1) is equal to 1/2, we have:
1/2 = x
2. Set up an integral to find the area:
∫(0 to 1/2) (x - 2^(-1)) dx
This integral represents the area between the curves y = f(x) and y = x, from x = 0 to x = 1/2.
3. Integrate the function:
∫(0 to 1/2) (x - 2^(-1)) dx = [x^2/2 - 2^(-1)x] (0 to 1/2)
Plug in the limits of integration:
[1/8 - 1/4] = -1/8
4. Take the absolute value of the result since the area cannot be negative:
| -1/8 | = 1/8
Therefore, the exact area of the region in the first quadrant bounded by the curves y = f(x) = 2^(-1) and y = x is 1/8.
B. Find the inverse function y = f^(-1)(x):
To find the inverse function of y = f(x) = 2^(-1), we need to swap the x and y variables and solve for y:
x = 2^(-1)
Swap x and y:
y = 2^(-1)
This means the inverse function is also y = f^(-1)(x) = 2^(-1).
C. Find the exact area of the region in the first quadrant bounded by the curves y = f^(-1)(x) and y = x:
Since the given function y = f(x) = 2^(-1) and its inverse y = f^(-1)(x) = 2^(-1) are symmetric about the line y = x, the area bounded by these two functions in the first quadrant will be the same as the area bounded by y = f(x) = 2^(-1) and y = x.
In other words, the exact area of the region in the first quadrant bounded by the curves y = f^(-1)(x) and y = x will also be 1/8. This is because the region is symmetric, and any points inside this region can be reflected across the line y = x to obtain corresponding points on the other side, maintaining the same area.
Therefore, the exact area of the region in the first quadrant bounded by the curves y = f^(-1)(x) and y = x is 1/8.
A. To find the exact area of the region in the first quadrant bounded by the curves y = 2^(-1) and y = x, we need to find the intersection points between the two curves.
Setting the two equations equal to each other, we have:
2^(-1) = x
Taking the logarithm (base 2) of both sides, we get:
-1 = log2(x)
Rewriting the equation in exponential form, we have:
2^(-1) = x
Using the fact that 2^(-1) is equal to 1/2, we can rewrite the equation as:
1/2 = x
So the two curves intersect at the point (1/2, 1/2).
To find the area, we need to integrate the difference between the two curves from x = 0 to x = 1/2:
A = ∫(2^(-1) - x) dx from 0 to 1/2
Integrating the expression, we have:
A = [(x - x^2)/2 - (x^2/2)] from 0 to 1/2
A = [(1/2 - 1/4)/2 - (1/4)/2] - [(0 - 0)/2 - (0/2)]
A = [(1/4)/2 - (1/4)/2] - [0 - 0]
A = 1/8 - 0
A = 1/8
Therefore, the exact area of the region in the first quadrant bounded by the curves y = 2^(-1) and y = x is 1/8.
B. To find the inverse function of y = 2^(-1), we need to switch the x and y variables and then solve for y:
x = 2^(-1)
Taking the logarithm (base 2) of both sides, we get:
log2(x) = -1
Rewriting the equation in exponential form, we have:
2^(-1) = x
So the inverse function is:
y = 2^(-1)
C. From part A, we found that the exact area of the region in the first quadrant bounded by the curves y = 2^(-1) and y = x is 1/8.
Since we know that the inverse function of y = 2^(-1) is y = 2^(-1), the area of the region in the first quadrant bounded by the curves y = 2^(-1) and y = x will be the same as the area of the region bounded by the curves y = x and y = 2^(-1).
Therefore, the exact area of the region in the first quadrant bounded by the curves y = 2^(-1) and y = x is also 1/8.