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iren2701 [21]
3 years ago
6

True or false (Picture provided)

Mathematics
2 answers:
STatiana [176]3 years ago
7 0

Answer:

False.

Step-by-step explanation:

We have given a function and its inverse:

f(x) = (x-3)³+4  and    f⁻¹(x) = -3+∛ (x+4)

We have to find is it inverse of function or not.

We have given a function:

f(x) = (x-3)³+4

let f(x) = y we get,

y = (x-3)³+4

Interchange the place of x and y we get,

x= (y-3)³+4

(y-3)³ = x-4

y-3 = ∛ x-4

y = ∛ x-4 +3

Replacing y to f⁻¹(x)  we get,

f⁻¹(x) = ∛ x-4 +3

The inverse of given function is f⁻¹(x) = ∛ x-4 +3.Which is different from the given inverse.

So, it is false.

aksik [14]3 years ago
3 0
<h2>Answer:</h2><h3>False</h3><h2>Step-by-step explanation:</h2>

Let's find the inverse function of f(x)=(x-3)^3+4 to know whether this function has an inverse function f^{-1}(x)=-3+\sqrt[3]{x+4}. So let's apply this steps:

a) Use the Horizontal Line Test to decide whether f has an inverse function.

Given that f(x) is a cubic function there is no any horizontal line that intersects the graph of f at more than one point. Thus, the function is one-to-one and has an inverse function.

b) Replace f(x) by y in the equation for f(x).

y=(x-3)^3+4

c) Interchange the roles of x and y and solve for y

x=(y-3)^3+4 \\ \\ \therefore x-4=(y-3)^3 \\ \\ \therefore (y-3)^3=x-4 \\ \\ \therefore y-3=\sqrt[3]{x-4} \\ \\ \therefore y=\sqrt[3]{x-4}+3

d) Replace y by f^{-1}(x) in the new equation.

f^{-1}(x)=\sqrt[3]{x-4}+3

So this is in fact the inverse function and it isn't the same given function. Therefore, the statement is false

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3 years ago
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A cylinder has a radius of 1 inch and a height of 1 inch. what is the approximate volume of the cylinder round to the nearest hu
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Step-by-step explanation:

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Hi there!

v = ±\sqrt{\frac{2K}{m} }

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K = 1/2mv²

Isolate for the variable "v". We can begin by dividing both sides by 1/2. (Multiply by the reciprocal, or 2):

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Continue isolating by dividing both sides by "m":

2K / m = v²

Take the square root of both sides. Remember that the solution can either be positive or negative since there are positive and negative roots.

v = ±\sqrt{\frac{2K}{m} }

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