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Tresset [83]
2 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]2 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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Step-by-step explanation:

Given

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1 year ago
What is the average rate of change for h(t) between t=0 and t=2
nordsb [41]

Part a: Option D 18 feet per second

Part b: increasing

Solution:

Height h(t)=-16 t^{2}+50 t+3

Part a: To find the average rate of change for h(t) between t = 0 and t = 2.

Substitute t = 0 in h(t).

h(0)=-16 (0)^{2}+50 (0)+3

h(0) = 3

Substitute t = 2 in h(t).

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Average rate of change formula:

           $=\frac{h(b)-h(a)}{b-a}

Here, a = 0 and b = 2.

           $=\frac{h(2)-h(0)}{2-0}

           $=\frac{39-3}{2}

           $=\frac{36}{2}

           = 18

Average rate of change = 18 feet per second

Option D is the correct answer.

Part b:

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sweet [91]

Answer:

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i added 30% and 40

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