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Tresset [83]
3 years ago
11

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

verse function.
Mathematics
1 answer:
Anon25 [30]3 years ago
8 0

We have been given a quadratic function f(x)=(x-5)^{2} +2 and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is [2,\infty).

If we restrict the domain of this function to either (-\infty,5] or [5,\infty), it will become one to one function.

Let us know find its inverse.

y=(x-5)^{2}+2

Upon interchanging x and y, we get:

x=(y-5)^{2}+2

Let us now solve this function for y.

(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\

Hence, the inverse function would be f^{-1}(x)=5+\sqrt{x-2} if we restrict the domain of original function to [5,\infty) and the inverse function would be f^{-1}(x)=5-\sqrt{x-2} if we restrict the domain to (-\infty,5].

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Answer:

(i) The area of the rabbit cage when the width is 5.2 m is 81.5 m²

(ii) The area of the rabbit cage if Wilson has 40 meters of wire mesh is 75 m²

Step-by-step explanation:

(i) The given relation of the area, A to the width P of the rabbit cage is A = 3·p²

The graph of the function between the values of 0 and 6 inclusive is found as follows;

A,              3·p²

0,               0

1,                1

2,               12

3,               27

4,               48

5,               75

6,               108

Please find attached the graph of A to 3·p²

From the graph, we have when the the width, p, of the rabbit cage = 5.2, the area, A ≈ 81.5 m²

The area of the rabbit cage when the width is 5.2 m = 81.5 m²

(ii) Also from the graph given that the total wire mess with Wilson = 40 meters, we have;

The formula for the perimeter of the cage = The formula for the perimeter of a rectangle = 2×length + 2×width

The formula for the perimeter of the cage = 2×3×p + 2× p = 8·p

Where the total length of the wire mesh available = 40 meters for the cage

The 40 meters of wire mesh will be used round the perimeter of the cage

∴ 40 m. = 8·p

p = 40/8 = 5 m.

At p = 5 m. the area is given as A = 75 m².

Therefore, the area of the rabbit cage if Wilson has 40 meters of wire mesh = 75 m².

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