Review the graph of function h(x). Which point is on the graph of the inverse function h–1(x)?
A. (2, 4)
B. (0, one-half)
C. (0, 1)
D. (–5, 1)
Answers
Answer:
C
Step-by-step explanation:
Point (0, 1) is on the graph of the inverse function h⁻¹(x).
What is an Inverse Function?If f(x) is an original function with output y, then we can get inverse of the original function by putting y in the inverse function to obtain x.The domain of original function becomes the range of its inverse function and the range of the original function becomes the domain of its inverse function.Given: The graph of function y = h(x).
According to the graph, the function is a straight line:
y-intercept is (c) = 2
Consider two points on the graph, y = h(x).
Let (x₁, y₁) = (0, 2) and (x₂, y₂) = (2, -2).
Now, the slope (m) of the function will be calculated as:
⇒ m = (y₂ - y₁) / (x₂ - x₁)
⇒ m = (-2 - 2) / (2 - 0)
⇒ m = -2
Now, with c = 2 and m = -2 the equation of a given line will be calculated as:
The general equation of line is:
⇒ y = mx + c
⇒ y = -2x + 2
∴ y = h(x) = -2x + 2 is the given equation of line.
Now to find the h⁻¹(x), we have to interchange the values of x and y from the function h(x) and then obtain the relation for the general equation of line.
h⁻¹(x) will be calculated as:
⇒ x = -2y + 2
⇒ x - 2 = -2y
⇒ y = (2 - x)/2
∴ y = h⁻¹(x) = (2 - x)/2
Now, we will find which point is on the graph of h⁻¹(x).
Option (A): (2, 4)
put, x = 2 in the equation y = (2 - x)/2
⇒ y = 0
⇒ (2, 0) ≠ (2, 4)
∴ (2, 4) is not the point on h⁻¹(x).
Option (B): (0, 1/2)
put x = 0 in the equation y = (2 - x)/2
⇒ y = 1
⇒ (0, 1) ≠ (0, 1/2)
∴ (0, 1/2) is not the point on h⁻¹(x).
Option (C): (0, 1)
put x = 0 in the equation y = (2 - x)/2
⇒ y = 1
⇒ (0, 1) = (0, 1)
∴ (0, 1) is the point on h⁻¹(x).
Option (D): (–5, 1)
put x = -5 in the equation y = (2 - x)/2
⇒ y = 3.5
⇒ (-5, 3.5) ≠ (–5, 1)
∴ (–5, 1) is not the point on h⁻¹(x).
Therefore, the point on the graph of the inverse function of h⁻¹(x) is (0, 1).
Learn about the Inverse Function here: https://brainly.com/question/16406473?referrer=searchResults
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Hey there! I'm happy to help!
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Have a wonderful day! :D
Answer:
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