Manipulations with exponents
If a and b are any real numbers, then:
x^a\cdot x^b=x^{a+b}
2^3\cdot 2^2=2^5
\left(x-1\right)^{\frac{1}{2}}\cdot \left(x-1\right)^{2}=\left(x-1\right)^{\frac{1}{2}+2}=\left(x-1\right)^{\frac{5}{2}}
If a and b are any real numbers, then:
\dfrac{x^a}{x^b}=x^{a-b}
\dfrac{2^3}{2^2}=2^{3-2}=2^1=2
\dfrac{\left(x-1\right)^{\frac{1}{2}}}{\left(x-1\right)^2}=\left(x-1\right)^{\frac{1}{2}-2}\\=\left(x-1\right)^{\frac{-3}{2}}\\\dfrac{1}{\left(x-1\right)^{\frac{3}{2}}}
In particular, if b is a real number, then:
\dfrac{1}{x^b}=\dfrac{x^0}{x^b}=x^{-b}
\dfrac{1}{x^3}=x^{-3}
If a and b are real numbers, then:
\left(x^a\right)^b=x^{ab}
\left(2^3\right)^2=2^6
\left(\left(x-1\right)^7\right)^{-\frac{1}{2}}=\left(x-1\right)^{-\frac{7}{2}}=\dfrac{1}{\left(x-1\right)^{\frac{7}{2}}}
Radical functions and rational exponents
Matching radical functions and radical exponents
Suppose a is a positive odd number and b a real number. Then there is a unique real number which satisfies the following equation:
x^a=b
That real number is denoted by b^{1/a} and agrees with the x -component of the intersection of the horizontal line y=b and the graph of the function y=x^a.
Consider the following equation:
a^3=27
Since the exponent is odd, 3 is the unique real number such that 3^3=27. We denote:
3=27^{1/3}
Suppose a is a positive even number and b a nonnegative number. Then there is a unique nonnegative real number which satisfies the following equation:
x^a=b
That number is denoted by b^{1/a}.
The negative number -b^{1/a} is also a solution to the equation. Both of these solutions to the equation correspond to the x -components of the intersections between the horizontal line y=b and the graph of the function y=x^a.
Consider the following equation:
a^2=16
Since the exponent is even, 4 is the unique nonnegative real number such that 4^2=16. We denote:
4=16^{1/2}
The equation also has a negative solution:
a=-16^{1/2}=-4
Radical function
A radical function is any function of the following form:
f\left(x\right)= \sqrt[n]{g\left(x\right)}=\left(g\left(x\right)\right)^{\frac{1}{n}}
Where g\left(x\right) is a polynomial function.
The following function is an example of a radical function:
f\left(x\right)=\sqrt[3]{x^2-2x+6}
Domain and range
Domain of a radical function
Suppose that f\left(x\right)=\left(g\left(x\right)\right)^{\frac{1}{n}} is a radical function. Then the domain of f\left(x\right) is one of the following:
- All real numbers if n is odd.
- All numbers x such that g\left(x\right)\geq 0 if n is even.
Consider the following rational function:
f\left(x\right)=\left(x^3-2x\right)^{\frac{1}{7}}
7 is an odd number. Therefore, the domain of f\left(x\right) is all real numbers.
Consider the following rational function:
f\left(x\right)=\left(x^2-4\right)^{\frac{1}{6}}
6 is an even number. Therefore, the domain of f\left(x\right) is all real numbers x satisfying the following:
x^2-4\geq 0
Observe that x^2-4=\left(x-2\right)\left(x+2\right) has x -intercepts at \left(-2{,}0\right) and \left(2{,}0\right). Therefore the domain of the rational function is:
\left(-\infty, -2\left]\cup \right[2,\infty\right)
Sometimes finding the range of a radical function is too difficult. However, there are some general guidelines.
Let n be an odd integer. Consider the following radical function:
f\left(x\right)=x^{1/n}
The range of f\left(x\right) is all real numbers.
Consider the following function:
f\left(x\right)=x^{1/5}
The range of f\left(x\right) is all real numbers since 5 is an odd number.
Let n be an even integer. Consider the following radical function:
f\left(x\right)=x^{1/n}
The range of f\left(x\right) is \left[0,\infty\right).
Consider the following function:
f\left(x\right)=x^{1/12}
The range of f\left(x\right) is all real numbers since 12 is an even number.
Let g\left(x\right) be a function and suppose the range of g\left(x\right) is \left[a,b\right] where a,b are real numbers. If n is an odd number, then the range of the function g\left(x\right)^{1/n} is \left[a^{1/n},b^{1/n}\right].
Consider the following function:
f\left(x\right)=\left(2\sin\left(x\right)\right)^{1/3}
The range of f\left(x\right) is \left[-2^{1/3},2^{1/3}\right] because the range of 2\sin\left(x\right) is \left[-2{,}2\right] and 3 is an odd number.
The same conclusion of the proposition also follows if we consider intervals of the form \left(a,b\right) where a and b are any real numbers and can possibly be infinite.
Consider the following function:
f\left(x\right)=\left(x^2+8\right)^{1/3}
Then the range of x^2+8 is \left[8,\infty\right) and therefore the range of f\left(x\right) is \left[2,\infty\right).
Growth, decay and graphical representation
Suppose n is an odd number. Then the graph of y=x^n passes the horizontal line test and y=x^{1/n} is its inverse function. Therefore, the graph of y=x^{1/n} is obtained by reflecting the graph of y=x^n across the line y=x.
The larger the value of n, the slower the rate which the graph of y=x^{1/n} grows towards infinity.
Suppose n is an even number. Then the graph of y=x^n passes the horizontal line test if we restrict the domain of the function to \left[0,\infty\right) and y=x^{1/n} is its inverse function. Therefore, the graph of y=x^{1/n} is obtained by reflecting the graph of y=x^n with a restricted domain of \left[0,\infty\right) across the line y=x.
The larger the value of n the slower the rate which the graph of y=x^{1/n} grows towards infinity.
Equations and inequalities with radical functions
Equations
Let n be an odd number and a a real number. Consider an equation of the following form:
\left(f\left(x\right)\right)^{\frac{1}{n}}=a
The equation is equivalent to solving the following equation:
f\left(x\right)=a^n
Consider the following equation:
\left(x^2-2x\right)^{1/3}=-1
The equation is equivalent to solving the following equation:
x^2-2x=\left(-1\right)^3
Or:
x^2-2x=-1
Solve it:
x^2-2x+1=0
\left(x-1\right)^2=0
x=1
1 is the only solution to the equation.
Let n be an even number and a a real number. Consider an equation of the following form:
\left(f\left(x\right)\right)^{\frac{1}{n}}=a
Every solution to the equation is a solution to the following equation:
f\left(x\right)=a^n.
However, not every solution to f\left(x\right)=a^n may be a solution to the original equation \left(f\left(x\right)\right)^{\frac{1}{n}}=a. To solve the equation \left(f\left(x\right)\right)^{\frac{1}{n}}=a :
- First solve the equation f\left(x\right)=a^n.
- Take all solutions to the equation f\left(x\right)=a^n and verify which are solutions to the equation \left(f\left(x\right)\right)^{\frac{1}{n}}=a by plugging in the potential solutions to the equation and see which satisfy the equation.
Consider the following equation:
\sqrt{x-4}=4-x
All the solutions of this equation are solutions of the following equation:
x-4=\left(4-x\right)^2
First solve this equation:
x-4=x^2-8x+16\\0=x^2-9x+20\\0=\left(x-5\right)\left(x-4\right)
The only possible solutions are x=5 and x=4.
Check if those numbers are solutions to the first equation. Evaluating the original equation with those two numbers yields the following:
\sqrt{5-4}=4-5
1=-1
And :
\sqrt{4-4}=4-4
0=0
Since plugging in 5 gave us 1=-1, the number x=5 is not a solution. However, x=4 is a valid solution since 0=0.
Consider the following equation:
\left(x^2+1\right)^{1/2}=\sqrt{10}
The equation is equivalent to:
x^2+1=10
The solutions to x^2=9 are x=3 and x=-3. Plugging 3 and -3 into the original equation:
- \left(3^2+1\right)^{1/2}=\left(9+1\right)^{\frac{1}{2}}=10^{\frac{1}{2}}=\sqrt{10}
- \left(\left(-3\right)^2+1\right)^{1/2}=\left(9+1\right)^{\frac{1}{2}}=10^{\frac{1}{2}}=\sqrt{10}
Therefore both, 3 and -3 are solutions to the original equation as well.
Inequalities
Let n be an odd number and consider an inequality of the following form:
\left(f\left(x\right)\right)^{\frac{1}{n}}\leq a
Solving this inequality is equivalent to solving the following inequality:
f\left(x\right)\leq a^n
Consider the following inequality:
\left(x+1\right)^{1/3}\leq 2
The inequality is equivalent to the following:
x+1\leq 2^3=8
x\leq 7
Therefore, the set of solutions to the original inequality is:
\left(-\infty,7\right].
Similar rules apply when handling an inequality with a radical expression and the quantifiers \geq, \lt, or \gt.
Let n be an even number and consider an inequality of the following form:
\left(f\left(x\right)\right)^{\frac{1}{n}}\leq a
The inequality is solved as follows:
- Find all solutions to the equation \left(f\left(x\right)\right)^{\frac{1}{n}}= a.
- Order the solutions found on a number line and select intermediate values between the intervals defined by the solutions.
- Plug in the intermediate values into the original inequality. If the inequality is satisfied for the intermediate value, then so is every other value in that interval.
Consider the following inequality:
\sqrt{x-4}\geq 4-x
The solution to the equation \sqrt{x-4}= 4-x is x=4.
To determine all solutions to the inequality, we must select a number less than 4 and a number greater than 4, and then plug those numbers into the inequality to see if they make the inequality true.
Plugging in 5 we arrive at 1\geq -1 which is true and plugging in x=3 we arrive at \sqrt{-1}\geq 1, which is not true.
Therefore, the solution set to the inequality is \left[4,\infty\right).
Consider the following inequality:
\sqrt{x^2+1}\lt \sqrt{10}
The solutions to the equation \sqrt{x^2+1} = \sqrt{10} are x=3 and x=-3.
To determine all solutions to the inequality, we must select a number less than -3, a number between -3 and 3 and a number greater than 3, and then plug those numbers into the inequality to see if they make the inequality true.
- Plugging in x=-4 we arrive at \sqrt{17}\lt \sqrt{10}, which is not true.
- Plugging in x=0 we arrive at 1\lt \sqrt{10}, which is true.
- Plugging in x=4, we arrive at \sqrt{17}\lt \sqrt{10} which is not true.
Therefore, the solution set to the inequality is \left(-3{,}3\right).
Similar rules apply when handling an inequality with a radical expression and the quantifiers \geq, \lt, or \gt.