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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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A Confidence interval is desired for the true average stray-load loss mu (watts) for a certain type of induction motor when the
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Answer:

A sample size of 35 is needed.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, find the width W as such

W = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large must the sample size be if the width of the 95% interval for mu is to be 1.0:

We need to find n for which W = 1.

We have that \sigma^{2} = 9, then \sigma = \sqrt{\sigma^{2}} = \sqrt{9} = 3. So

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