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Use Radicals in Functions

Evaluate a radical function

Evaluate a Radical Function

In this section we will extend our previous work with functions to include radicals. If a function is defined by a radical expression, we call it a radical function.

The square root function is \(f(x)=\sqrt[]{x}.\)

The cube root function is \(f(x)=\sqrt[3]{x}.\)

To evaluate a radical function, we find the value of f(x) for a given value of x just as we did in our previous work with functions.

Example

Try it.

For the function \(f(x)=\sqrt{2x-1},\) find ⓐ \(f(5)\) ⓑ \(f(-2).\)

Solution

ⓐ

	\(f(x)=\sqrt{2x-1}\)
To evaluate \(f(5),\) substitute 5 for \(x.\)	\(f(5)=\sqrt{2\cdot 5-1}\)
Simplify.	\(f(5)=\sqrt{9}\)
Take the square root.	\(f(5)=3\)

ⓑ

	\(\ f(x)=\sqrt{2x-1}\)
To evaluate \(f(-2),\) substitute −2 for \(x.\)	\(\ f(-2)=\sqrt{2(-2)-1}\)
Simplify.	\(\ f(-2)=\sqrt{-5}\)

Since the square root of a negative number is not a real number, the function does not have a value at \(x=-2.\)

We follow the same procedure to evaluate cube roots.

Example

Try it.

For the function \(g(x)=\sqrt[3]{x-6},\) find ⓐ \(g(14)\) ⓑ \(g(-2).\)

Solution

ⓐ

	\(\ g(x)=\sqrt[3]{x-6}\)
To evaluate \(g(14),\) substitute 14 for \(x.\)	\(\ g(14)=\sqrt[3]{14-6}\)
Simplify.	\(\ g(14)=\sqrt[3]{8}\)
Take the cube root.	\(\ g(14)=2\)

ⓑ

	\(\ g(x)=\sqrt[3]{x-6}\)
To evaluate \(g(-2),\) substitute −2 for \(x.\)	\(\ g(-2)=\sqrt[3]{-2-6}\)
Simplify.	\(\ g(-2)=\sqrt[3]{-8}\)
Take the cube root.	\(\ g(-2)=-2\)

The next example has fourth roots.

Example

Try it.

For the function \(f(x)=\sqrt[4]{5x-4},\) find ⓐ \(f(4)\) ⓑ \(f(-12)\)

Solution

ⓐ

	\(\ f(x)=\sqrt[4]{5x-4}\)
To evaluate \(f(4),\) substitute 4 for \(x.\)	\(\ f(4)=\sqrt[4]{5\cdot 4-4}\)
Simplify.	\(\ f(4)=\sqrt[4]{16}\)
Take the fourth root.	\(\ f(4)=2\)

ⓑ

	\(\ f(x)=\sqrt[4]{5x-4}\)
To evaluate \(f(-12),\) substitute −12 for \(x.\)	\(\ f(-12)=\sqrt[4]{5(-12)-4}\)
Simplify.	\(\ f(-12)=\sqrt[4]{-64}\)

Since the fourth root of a negative number is not a real number, the function does not have a value at \(x=-12.\)

Find the Domain of a Radical Function

To find the domain and range of radical functions, we use our properties of radicals. For a radical with an even index, we said the radicand had to be greater than or equal to zero as even roots of negative numbers are not real numbers. For an odd index, the radicand can be any real number. We restate the properties here for reference.

So, to find the domain of a radical function with even index, we set the radicand to be greater than or equal to zero. For an odd index radical, the radicand can be any real number.

Example

Try it.

Find the domain of the function, \(f(x)=\sqrt{3x-4}.\) Write the domain in interval notation.

Solution

Since the function, \(f(x)=\sqrt{3x-4}\) has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0. We set the radicand to be greater than or equal to 0 and then solve to find the domain.

	\(3x-4\ge 0\)
Solve.	\(3x\ge 4\)
	\(x\ge \frac{4}{3}\)

The domain of \(f(x)=\sqrt{3x-4}\) is all values \(x\ge \frac{4}{3}\) and we write it in interval notation as \([\frac{4}{3},\infty ).\)

Example

Try it.

Find the domain of the function, \(g(x)=\sqrt{\frac{6}{x-1}}.\) Write the domain in interval notation.

Solution

Since the function, \(g(x)=\sqrt{\frac{6}{x-1}}\) has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0.

The radicand cannot be zero since the numerator is not zero.

For \(\frac{6}{x-1}\) to be greater than zero, the denominator must be positive since the numerator is positive. We know a positive divided by a positive is positive.

We set \(x-1>0\) and solve.

Solve.

\(\begin{array}{lll}x-1 & > & 0 \\ x & > & 1\end{array}\)

Also, since the radicand is a fraction, we must realize that the denominator cannot be zero.

We solve \(x-1=0\) to find the value that must be eliminated from the domain.

Solve.

\(\begin{array}{lll}x-1 & = & 0 \\ x & = & 1\ \text{so}\ x\ne 1\ \text{in the domain.}\end{array}\)

Putting this together we get the domain is \(x>1\) and we write it as \((1,\infty ).\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Graph Radical Functions

Before we graph any radical function, we first find the domain of the function. For the function, \(f(x)=\sqrt{x},\) the index is even, and so the radicand must be greater than or equal to 0.

This tells us the domain is \(x\ge 0\) and we write this in interval notation as \([0,\infty ).\)

Previously we used point plotting to graph the function, \(f(x)=\sqrt{x}.\) We chose x-values, substituted them in and then created a chart. Notice we chose points that are perfect squares in order to make taking the square root easier.

Once we see the graph, we can find the range of the function. The y-values of the function are greater than or equal to zero. The range then is \([0,\infty ).\)

Example

Try it.

For the function \(f(x)=\sqrt{x+3},\)

ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ Since the radical has index 2, we know the radicand must be greater than or equal to zero. If \(x+3\ge 0,\) then \(x\ge -3.\) This tells us the domain is all values \(x\ge -3\) and written in interval notation as \([-3,\infty ).\)

ⓑ To graph the function, we choose points in the interval \([-3,\infty )\) that will also give us a radicand which will be easy to take the square root.

In our previous work graphing functions, we graphed \(f(x)={x}^{3}\) but we did not graph the function \(f(x)=\sqrt[3]{x}.\) We will do this now in the next example.

Example

Try it.

For the function \(f(x)=\sqrt[3]{x},\) ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

Solution

ⓐ Since the radical has index 3, we know the radicand can be any real number. This tells us the domain is all real numbers and written in interval notation as \((\text{-}\infty ,\infty )\)

ⓑ To graph the function, we choose points in the interval \((\text{-}\infty ,\infty )\) that will also give us a radicand which will be easy to take the cube root.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Properties of \(\sqrt[n]{a}\)
- When n is an even number and:
  \(a\ge 0,\) then \(\sqrt[n]{a}\) is a real number.
  \(a<0,\) then \(\sqrt[n]{a}\) is not a real number.
- When n is an odd number, \(\sqrt[n]{a}\) is a real number for all values of a.
Domain of a Radical Function
- When the index of the radical is even, the radicand must be greater than or equal to zero.
- When the index of the radical is odd, the radicand can be any real number.

Use Radicals in Functions

Evaluate a Radical Function

In the following exercises, evaluate each function.

Try it.

\(f(x)=\sqrt{4x-4},\) find ⓐ \(f(5)\) ⓑ \(f(0).\)

Solution

ⓐ \(f(5)=4\) ⓑ no value at \(x=0\)

Try it.

\(f(x)=\sqrt{6x-5},\) find ⓐ \(f(5)\) ⓑ \(f(-1).\)

Try it.

\(g(x)=\sqrt{6x+1},\) find ⓐ \(g(4)\) ⓑ \(g(8).\)

Solution

ⓐ \(g(4)=5\) ⓑ \(g(8)=7\)

Try it.

\(g(x)=\sqrt{3x+1},\) find ⓐ \(g(8)\) ⓑ \(g(5).\)

Try it.

\(F(x)=\sqrt{3-2x},\) find ⓐ \(F(1)\) ⓑ \(F(-11).\)

Solution

ⓐ \(F(1)=1\) ⓑ \(F(-11)=5\)

Try it.

\(F(x)=\sqrt{8-4x},\) find ⓐ \(F(1)\) ⓑ \(F(-2).\)

Try it.

\(G(x)=\sqrt{5x-1},\) find ⓐ \(G(5)\) ⓑ \(G(2).\)

Solution

ⓐ \(G(5)=2\sqrt{6}\) ⓑ \(G(2)=3\)

Try it.

\(G(x)=\sqrt{4x+1},\) find ⓐ \(G(11)\) ⓑ \(G(2).\)

Try it.

\(g(x)=\sqrt[3]{2x-4},\) find ⓐ \(g(6)\) ⓑ \(g(-2).\)

Solution

ⓐ \(g(6)=2\) ⓑ \(g(-2)=-2\)

Try it.

\(g(x)=\sqrt[3]{7x-1},\) find ⓐ \(g(4)\) ⓑ \(g(-1).\)

Try it.

\(h(x)=\sqrt[3]{{x}^{2}-4},\) find ⓐ \(h(-2)\) ⓑ \(h(6).\)

Solution

ⓐ \(h(-2)=0\) ⓑ \(h(6)=2\sqrt[3]{4}\)

Try it.

\(h(x)=\sqrt[3]{{x}^{2}+4},\) find ⓐ \(h(-2)\) ⓑ \(h(6).\)

Try it.

For the function \(f(x)=\sqrt[4]{2{x}^{3}},\) find ⓐ \(f(0)\) ⓑ \(f(2).\)

Solution

ⓐ \(f(0)=0\) ⓑ \(f(2)=2\)

Try it.

For the function \(f(x)=\sqrt[4]{3{x}^{3}},\) find ⓐ \(f(0)\) ⓑ \(f(3).\)

Try it.

For the function \(g(x)=\sqrt[4]{4-4x},\) find ⓐ \(g(1)\) ⓑ \(g(-3).\)

Solution

ⓐ \(g(1)=0\) ⓑ \(g(-3)=2\)

Try it.

For the function \(g(x)=\sqrt[4]{8-4x},\) find ⓐ \(g(-6)\) ⓑ \(g(2).\)

Find the Domain of a Radical Function

In the following exercises, find the domain of the function and write the domain in interval notation.

Try it.

\(f(x)=\sqrt{3x-1}\)

Solution

\([\frac{1}{3},\infty )\)

Try it.

\(f(x)=\sqrt{4x-2}\)

Try it.

\(g(x)=\sqrt{2-3x}\)

Solution

\((\text{-}\infty ,\frac{2}{3}]\)

Try it.

\(g(x)=\sqrt{8-x}\)

Try it.

\(h(x)=\sqrt{\frac{5}{x-2}}\)

Solution

\((2,\infty )\)

Try it.

\(h(x)=\sqrt{\frac{6}{x+3}}\)

Try it.

\(f(x)=\sqrt{\frac{x+3}{x-2}}\)

Solution

\((\text{-}\infty ,-3]\cup (2,\infty )\)

Try it.

\(f(x)=\sqrt{\frac{x-1}{x+4}}\)

Try it.

\(g(x)=\sqrt[3]{8x-1}\)

Solution

\((\text{-}\infty ,\infty )\)

Try it.

\(g(x)=\sqrt[3]{6x+5}\)

Try it.

\(f(x)=\sqrt[3]{4{x}^{2}-16}\)

Solution

\((\text{-}\infty ,\infty )\)

Try it.

\(f(x)=\sqrt[3]{6{x}^{2}-25}\)

Try it.

\(F(x)=\sqrt[4]{8x+3}\)

Solution

\([-\frac{3}{8},\infty )\)

Try it.

\(F(x)=\sqrt[4]{10-7x}\)

Try it.

\(G(x)=\sqrt[5]{2x-1}\)

Solution

\((\text{-}\infty ,\infty )\)

Try it.

\(G(x)=\sqrt[5]{6x-3}\)

Graph Radical Functions

In the following exercises, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.

Try it.

\(f(x)=\sqrt{x+1}\)

Solution

ⓐ domain: \([-1,\infty )\)
ⓑ

ⓒ \([0,\infty )\)

Try it.

\(f(x)=\sqrt{x-1}\)

Try it.

\(g(x)=\sqrt{x+4}\)

Solution

ⓐ domain: \([-4,\infty )\)
ⓑ

ⓒ \([0,\infty )\)

Try it.

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

Try it.

\(f(x)=\sqrt{x}+2\)

Solution

ⓐ domain: \([0,\infty )\)
ⓑ

ⓒ \([2,\infty )\)

Try it.

\(f(x)=\sqrt{x}-2\)

Try it.

\(g(x)=2\sqrt{x}\)

Solution

Try it.

\(g(x)=3\sqrt{x}\)

Try it.

\(f(x)=\sqrt{3-x}\)

Solution

Try it.

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

Try it.

\(g(x)=\text{-}\sqrt{x}\)

Solution

Try it.

\(g(x)=\text{-}\sqrt{x}+1\)

Try it.

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

Solution

Try it.

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

Try it.

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

Solution

Try it.

\(g(x)=\sqrt[3]{x-2}\)

Try it.

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

Solution

Try it.

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

Try it.

\(g(x)=\sqrt[3]{x}\)

Solution

Try it.

\(g(x)=\text{-}\sqrt[3]{x}\)

Try it.

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

Solution

Try it.

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

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve: \(1-2x\ge 0.\)
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

\((\text{-}\infty ,\frac{1}{2}]\)
For \(f(x)=3x-4,\) evaluate \(f(2),f(-1),f(0).\)
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

\(f(2)=2,f(-1)=-7,f(0)=-4\)
Graph \(f(x)=\sqrt{x}.\) State the domain and range of the function in interval notation.
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

\(\text{domain:}\ [0,\infty );\ \text{range:}\ [0,\infty )\)
For the function \(f(x)=\sqrt{2x-1},\) find ⓐ \(f(5)\) ⓑ \(f(-2).\)

जवाफ प्रकट गर्नुहोस्

ⓐ

\(f(x)=\sqrt{2x-1}\)
To evaluate \(f(5),\) substitute 5 for \(x.\) \(f(5)=\sqrt{2\cdot 5-1}\)
Simplify. \(f(5)=\sqrt{9}\)
Take the square root. \(f(5)=3\)
ⓑ

\(\ f(x)=\sqrt{2x-1}\)
To evaluate \(f(-2),\) substitute −2 for \(x.\) \(\ f(-2)=\sqrt{2(-2)-1}\)
Simplify. \(\ f(-2)=\sqrt{-5}\)

Since the square root of a negative number is not a real number, the function does not have a value at \(x=-2.\)
For the function \(f(x)=\sqrt{3x-2},\) find ⓐ \(f(6)\) ⓑ \(f(0).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(f(6)=4\) ⓑ no value at \(x=0\)
For the function \(g(x)=\sqrt{5x+5},\) find ⓐ \(g(4)\) ⓑ \(g(-3).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(g(4)=5\) ⓑ no value at \(f(-3)\)
For the function \(g(x)=\sqrt[3]{x-6},\) find ⓐ \(g(14)\) ⓑ \(g(-2).\)

जवाफ प्रकट गर्नुहोस्

ⓐ

\(\ g(x)=\sqrt[3]{x-6}\)
To evaluate \(g(14),\) substitute 14 for \(x.\) \(\ g(14)=\sqrt[3]{14-6}\)
Simplify. \(\ g(14)=\sqrt[3]{8}\)
Take the cube root. \(\ g(14)=2\)
ⓑ

\(\ g(x)=\sqrt[3]{x-6}\)
To evaluate \(g(-2),\) substitute −2 for \(x.\) \(\ g(-2)=\sqrt[3]{-2-6}\)
Simplify. \(\ g(-2)=\sqrt[3]{-8}\)
Take the cube root. \(\ g(-2)=-2\)
For the function \(g(x)=\sqrt[3]{3x-4},\) find ⓐ \(g(4)\) ⓑ \(g(1).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(g(4)=2\) ⓑ \(g(1)=-1\)
For the function \(h(x)=\sqrt[3]{5x-2},\) find ⓐ \(h(2)\) ⓑ \(h(-5).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(h(2)=2\)
ⓑ \(h(-5)=-3\)
For the function \(f(x)=\sqrt[4]{5x-4},\) find ⓐ \(f(4)\) ⓑ \(f(-12)\)

जवाफ प्रकट गर्नुहोस्

ⓐ

\(\ f(x)=\sqrt[4]{5x-4}\)
To evaluate \(f(4),\) substitute 4 for \(x.\) \(\ f(4)=\sqrt[4]{5\cdot 4-4}\)
Simplify. \(\ f(4)=\sqrt[4]{16}\)
Take the fourth root. \(\ f(4)=2\)
ⓑ

\(\ f(x)=\sqrt[4]{5x-4}\)
To evaluate \(f(-12),\) substitute −12 for \(x.\) \(\ f(-12)=\sqrt[4]{5(-12)-4}\)
Simplify. \(\ f(-12)=\sqrt[4]{-64}\)

Since the fourth root of a negative number is not a real number, the function does not have a value at \(x=-12.\)
For the function \(f(x)=\sqrt[4]{3x+4},\) find ⓐ \(f(4)\) ⓑ \(f(-1).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(f(4)=2\) ⓑ \(f(-1)=1\)
For the function \(g(x)=\sqrt[4]{5x+1},\) find ⓐ \(g(16)\) ⓑ \(g(3).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(g(16)=3\) ⓑ \(g(3)=2\)
Find the domain of the function, \(f(x)=\sqrt{3x-4}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

Since the function, \(f(x)=\sqrt{3x-4}\) has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0. We set the radicand to be greater than or equal to 0 and then solve to find the domain.
\(3x-4\ge 0\)
Solve. \(3x\ge 4\)
\(x\ge \frac{4}{3}\)
The domain of \(f(x)=\sqrt{3x-4}\) is all values \(x\ge \frac{4}{3}\) and we write it in interval notation as \([\frac{4}{3},\infty ).\)
Find the domain of the function, \(f(x)=\sqrt{6x-5}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

\([\frac{5}{6},\infty )\)
Find the domain of the function, \(f(x)=\sqrt{4-5x}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

\((\text{-}\infty ,\frac{4}{5}]\)
Find the domain of the function, \(g(x)=\sqrt{\frac{6}{x-1}}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

Since the function, \(g(x)=\sqrt{\frac{6}{x-1}}\) has a radical with an index of 2, which is even, we know the radicand must be greater than or equal to 0.
The radicand cannot be zero since the numerator is not zero.
For \(\frac{6}{x-1}\) to be greater than zero, the denominator must be positive since the numerator is positive. We know a positive divided by a positive is positive.
We set \(x-1>0\) and solve.

Solve. \(\begin{array}{lll}x-1 & > & 0 \\ x & > & 1\end{array}\)

Also, since the radicand is a fraction, we must realize that the denominator cannot be zero.
We solve \(x-1=0\) to find the value that must be eliminated from the domain.

Solve. \(\begin{array}{lll}x-1 & = & 0 \\ x & = & 1\ \text{so}\ x\ne 1\ \text{in the domain.}\end{array}\)

Putting this together we get the domain is \(x>1\) and we write it as \((1,\infty ).\)
Find the domain of the function, \(f(x)=\sqrt{\frac{4}{x+3}}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

\((-3,\infty )\)
Find the domain of the function, \(h(x)=\sqrt{\frac{9}{x-5}}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

\((5,\infty )\)
Find the domain of the function, \(f(x)=\sqrt[3]{2{x}^{2}+3}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

Since the function, \(f(x)=\sqrt[3]{2{x}^{2}+3}\) has a radical with an index of 3, which is odd, we know the radicand can be any real number. This tells us the domain is any real number. In interval notation, we write \((\text{-}\infty ,\infty ).\)
The domain of \(f(x)=\sqrt[3]{2{x}^{2}+3}\) is all real numbers and we write it in interval notation as \((\text{-}\infty ,\infty ).\)
Find the domain of the function, \(f(x)=\sqrt[3]{3{x}^{2}-1}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

\((\text{-}\infty ,\infty )\)
Find the domain of the function, \(g(x)=\sqrt[3]{5x-4}.\) Write the domain in interval notation.

जवाफ प्रकट गर्नुहोस्

\((\text{-}\infty ,\infty )\)
For the function \(f(x)=\sqrt{x+3},\)
ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

जवाफ प्रकट गर्नुहोस्

ⓐ Since the radical has index 2, we know the radicand must be greater than or equal to zero. If \(x+3\ge 0,\) then \(x\ge -3.\) This tells us the domain is all values \(x\ge -3\) and written in interval notation as \([-3,\infty ).\)
ⓑ To graph the function, we choose points in the interval \([-3,\infty )\) that will also give us a radicand which will be easy to take the square root.
ⓒ Looking at the graph, we see the y-values of the function are greater than or equal to zero. The range then is \([0,\infty ).\)
For the function \(f(x)=\sqrt{x+2},\) ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

जवाफ प्रकट गर्नुहोस्

ⓐ domain: \([-2,\infty )\)
ⓑ

ⓒ range: \([0,\infty )\)
For the function \(f(x)=\sqrt{x-2},\) ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

जवाफ प्रकट गर्नुहोस्

ⓐ domain: \([2,\infty )\)
ⓑ

ⓒ range: \([0,\infty )\)
For the function \(f(x)=\sqrt[3]{x},\) ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

जवाफ प्रकट गर्नुहोस्

ⓐ Since the radical has index 3, we know the radicand can be any real number. This tells us the domain is all real numbers and written in interval notation as \((\text{-}\infty ,\infty )\)
ⓑ To graph the function, we choose points in the interval \((\text{-}\infty ,\infty )\) that will also give us a radicand which will be easy to take the cube root.

ⓒ Looking at the graph, we see the y-values of the function are all real numbers. The range then is \((\text{-}\infty ,\infty ).\)
For the function \(f(x)=\text{-}\sqrt[3]{x},\)
ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

जवाफ प्रकट गर्नुहोस्

ⓐ domain: \(\text{(}-\infty ,\infty )\)
ⓑ

ⓒ range: \(\text{(}-\infty ,\infty )\)
For the function \(f(x)=\sqrt[3]{x-2},\)
ⓐ find the domain ⓑ graph the function ⓒ use the graph to determine the range.

जवाफ प्रकट गर्नुहोस्

ⓐ domain: \(\text{(}-\infty ,\infty )\)
ⓑ

ⓒ range: \(\text{(}-\infty ,\infty )\)
\(f(x)=\sqrt{4x-4},\) find ⓐ \(f(5)\) ⓑ \(f(0).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(f(5)=4\) ⓑ no value at \(x=0\)
\(f(x)=\sqrt{6x-5},\) find ⓐ \(f(5)\) ⓑ \(f(-1).\)
\(g(x)=\sqrt{6x+1},\) find ⓐ \(g(4)\) ⓑ \(g(8).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(g(4)=5\) ⓑ \(g(8)=7\)
\(g(x)=\sqrt{3x+1},\) find ⓐ \(g(8)\) ⓑ \(g(5).\)
\(F(x)=\sqrt{3-2x},\) find ⓐ \(F(1)\) ⓑ \(F(-11).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(F(1)=1\) ⓑ \(F(-11)=5\)
\(F(x)=\sqrt{8-4x},\) find ⓐ \(F(1)\) ⓑ \(F(-2).\)
\(G(x)=\sqrt{5x-1},\) find ⓐ \(G(5)\) ⓑ \(G(2).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(G(5)=2\sqrt{6}\) ⓑ \(G(2)=3\)
\(G(x)=\sqrt{4x+1},\) find ⓐ \(G(11)\) ⓑ \(G(2).\)
\(g(x)=\sqrt[3]{2x-4},\) find ⓐ \(g(6)\) ⓑ \(g(-2).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(g(6)=2\) ⓑ \(g(-2)=-2\)
\(g(x)=\sqrt[3]{7x-1},\) find ⓐ \(g(4)\) ⓑ \(g(-1).\)
\(h(x)=\sqrt[3]{{x}^{2}-4},\) find ⓐ \(h(-2)\) ⓑ \(h(6).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(h(-2)=0\) ⓑ \(h(6)=2\sqrt[3]{4}\)
\(h(x)=\sqrt[3]{{x}^{2}+4},\) find ⓐ \(h(-2)\) ⓑ \(h(6).\)
For the function \(f(x)=\sqrt[4]{2{x}^{3}},\) find ⓐ \(f(0)\) ⓑ \(f(2).\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(f(0)=0\) ⓑ \(f(2)=2\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Use Radicals in Functions

Evaluate a radical function
Find the domain of a radical function
Graph radical functions
When
When
When the
When the

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

तपाईँको आफ्नै प्रयास गर्नुहोस्

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

यसमा थप Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value