MHB What is the domain for ax^(1/3) + b?

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The discussion centers on determining the domain of the function ax^(1/3) + b. Initially, it is suggested that the domain is x ≥ 0 due to the presence of a radical. However, it is clarified that the cube root function is defined for all real numbers, meaning the domain encompasses all real numbers. This distinction is made to highlight that unlike square roots, cube roots can accept negative inputs. Ultimately, the correct domain for the function is confirmed to be all real numbers.
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Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 3b.

Specify the domain.

ax^(1/3) + b

Solution:

x^(1/3) means the cube root of x.

Since there is a radical here, I will say the domain is the radicand > or = 0.

So, x > or = 0.

Yes?
 
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RTCNTC said:
Precalculus by David Cohen, 3rd Edition
Chapter 1, Section 1.3.
Question 3b.

Specify the domain.

ax^(1/3) + b

Solution:

x^(1/3) means the cube root of x.

Since there is a radical here, I will say the domain is the radicand > or = 0.

So, x > or = 0.

Yes?

as cube root is defined for all real number (-ve number also) so the domain is set of real numbers
 
You are right. I found the following definition online:

"The domain of a cube root function is the set of all real numbers. Unlike a square root function which is limited to nonnegative numbers, a cube root can use all real numbers because it is possible for three negatives to equal a negative."
 

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