We have been given a quadratic function
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
.
If we restrict the domain of this function to either
or
, it will become one to one function.
Let us know find its inverse.
![y=(x-5)^{2}+2](https://tex.z-dn.net/?f=y%3D%28x-5%29%5E%7B2%7D%2B2)
Upon interchanging x and y, we get:
![x=(y-5)^{2}+2](https://tex.z-dn.net/?f=x%3D%28y-5%29%5E%7B2%7D%2B2)
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}\\](https://tex.z-dn.net/?f=%28y-5%29%5E%7B2%7D%3Dx-2%5C%5C%0Ay-5%3D%5Cpm%20%5Csqrt%7Bx-2%7D%5C%5C%0Ay%3D5%5Cpm%20%5Csqrt%7Bx-2%7D%5C%5C)
Hence, the inverse function would be
if we restrict the domain of original function to
and the inverse function would be
if we restrict the domain to
.
Answer:
the answer is c
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Step-by-step explanation:
There are many ways, but here is one:
x = 100
Since ‘x’ is 100 and we want 100 more, we just need to add ‘x’ to 623
Answer: x + 623 (Make sure you define your ‘x’ value)
Answer:
D: 101, 135, 131, 99, 138, 136, 140
Step-by-step explanation:
edg2020 (do not get this confused with the question: "which set contains no outliers"