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What are the domain and range of the function?

What are the domain and range of the function?

jbouyer  Mar 22, 2018

Best Answer 

 #1
avatar+2248 
+2

The domain is the set of all x-values that outputs a real y-value for a given equation.

 

\(f(x)=\sqrt[3]{x}+2\)

 

In this case, this particular function will accept any input and produce a real output. Because this is a cube-root function, the cube-root can produce any result of reals, so \(x\in \mathbb{R}\text{ or }(-\infty,+\infty)\) . Only one answer choice fits this criterion, so the 3rd answer must be the correct one. Let's consider the next function!

 

\(f(x)=-4\sqrt{x}\)

 

When a variable is placed underneath the square root (known as the radicand), there is a certain domain restriction that must be taken into account: The radicand must be nonnegative; inserting a negative number as the radicand will result in a non-real output since no real number multiplied by itself can equal a negative number. Because of this fact, the domain is \(x\geq 0\text{ or }[0,+\infty)\) . Unfortunately, this is not enough to eliminate every answer choice because two answer choices include this as a possibility. Therefore, we will have to figure out the range.

 

The range is the set of all outputs that the function can produce

 

Because of the definition of the square root operation, it is implied that this is a case of the principal root. Disregarding the rest of the function, \(\sqrt{x}\) will always output a positive value or zero. Multiplying any positive number or zero by a negative number will always result in a negative number or zero, so the output is any real non-positive number or \(f(x)\leq 0\text{ or }(-\infty,0]\)

 

With this knowledge, only the third answer choice remains.

TheXSquaredFactor  Mar 22, 2018
 #1
avatar+2248 
+2
Best Answer

The domain is the set of all x-values that outputs a real y-value for a given equation.

 

\(f(x)=\sqrt[3]{x}+2\)

 

In this case, this particular function will accept any input and produce a real output. Because this is a cube-root function, the cube-root can produce any result of reals, so \(x\in \mathbb{R}\text{ or }(-\infty,+\infty)\) . Only one answer choice fits this criterion, so the 3rd answer must be the correct one. Let's consider the next function!

 

\(f(x)=-4\sqrt{x}\)

 

When a variable is placed underneath the square root (known as the radicand), there is a certain domain restriction that must be taken into account: The radicand must be nonnegative; inserting a negative number as the radicand will result in a non-real output since no real number multiplied by itself can equal a negative number. Because of this fact, the domain is \(x\geq 0\text{ or }[0,+\infty)\) . Unfortunately, this is not enough to eliminate every answer choice because two answer choices include this as a possibility. Therefore, we will have to figure out the range.

 

The range is the set of all outputs that the function can produce

 

Because of the definition of the square root operation, it is implied that this is a case of the principal root. Disregarding the rest of the function, \(\sqrt{x}\) will always output a positive value or zero. Multiplying any positive number or zero by a negative number will always result in a negative number or zero, so the output is any real non-positive number or \(f(x)\leq 0\text{ or }(-\infty,0]\)

 

With this knowledge, only the third answer choice remains.

TheXSquaredFactor  Mar 22, 2018

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