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Stage 4 - Stage 5

# Evaluating Cubic and Other Power Functions

Lesson

In mathematics, a power function is a function of the form $y=a\times x^r$y=a×xr where $a$a and $r$r are real numbers.

The coefficient $a$a is taken as non-zero.

So for example, $y=2x^3$y=2x3$f\left(x\right)=\sqrt{x}$f(x)=x$y=\sqrt{2}x^5$y=2x5$g\left(x\right)=-\frac{3}{x}$g(x)=3x and $y=-x^{-3}$y=x3 are all examples of power functions.

The power functions are quite comprehensive in scope, covering simple rectangular hyperbolae, basic polynomial functions including the cubic function, square root and cube root functions, etc.

## Issues when evaluating f(x) where x is negative

Suppose we consider the power function $f\left(x\right)=x^{1.2}$f(x)=x1.2. With technology, it is easy to verify that $f(2)=2^{1.2}\approx2.2974$f(2)=21.22.2974

However when you try $f(-2)=(-2)^{1.2}$f(2)=(2)1.2, some of the technology can show an incorrect 'error' response.

It is incorrect because $y=x^{\frac{6}{5}}$y=x65 is defined for all real $x$x irrespective of whether it's negative, positive or $0$0 . The reason for this, however, is quite technical.

Without going into too much detail, for certain types of exponents $r$r, the power function $y=ax^r$y=axr is defined for negative $x$x values.  Specifically it can be defined when the exponent $r$r can be expressed as a rational number $\frac{p}{q}$pq where $p$p and $q$q are integers in their lowest terms and $q$q is odd.

The last part of the last sentence is critical - in lowest terms with $q$q odd!

This is how we would correctly evaluate something like $f\left(-2\right)=\left(-2\right)^{1.2}$f(2)=(2)1.2.

 $f\left(-2\right)$f(−2) $=$= $\left(-2\right)^{1.2}$(−2)1.2 $=$= $\left(-1\right)^{\frac{6}{5}}\left(2\right)^{\frac{6}{5}}$(−1)65​(2)65​ $=$= $\left(\left(-1\right)^{\frac{1}{5}}\right)^6\left(2\right)^{\frac{6}{5}}$((−1)15​)6(2)65​ $=$= $\left(-1\right)^6\left(2\right)^{\frac{6}{5}}$(−1)6(2)65​ $=$= $1\times\left(2\right)^{\frac{6}{5}}$1×(2)65​ $=$= $2.2974...$2.2974...

Note that the term $\left(-1\right)^{\frac{1}{5}}$(1)15 posed no problems, but the same would not have been true for, say, a function like $f\left(x\right)=x^{\frac{5}{6}}$f(x)=x56 with the numbers in the exponent switched. In this case evaluating $f\left(-2\right)$f(2) would have been problematic.

## Odd and Even Power functions

All power functions of the form $f\left(x\right)=ax^r$f(x)=axr where $\left|r\right|$|r| is an odd number, are odd functions. Recall that an odd function $f$f is a function where $f\left(-x\right)=-f\left(x\right)$f(x)=f(x) for all $x$x in the natural domain of the function.

For example the power function $f\left(x\right)=3x^{-1}=\frac{3}{x}$f(x)=3x1=3x is odd because for all $x$x in the natural domain (note that the domain in this case is $x:x\in\Re,x\ne0$x:x,x0) we have:

$f\left(-x\right)=\frac{3}{-x}=-\frac{3}{x}=-f\left(x\right)$f(x)=3x=3x=f(x)

Odd functions possess $180^\circ$180°  rotational symmetry about the origin. Essentially this means that if you imagine rotating the curve anticlockwise about the origin $180^\circ$180°, it will fall back onto itself.

All power functions of the form $f\left(x\right)=ax^r$f(x)=axr where $\left|r\right|$|r| is an even number, are even functions. Recall that an even function $f$f is a function where $f\left(-x\right)=f\left(x\right)$f(x)=f(x) for all x in the natural domain of the function.

For example the power function $f\left(x\right)=3x^2$f(x)=3x2 is even because for all $x$x in the natural domain we have:

$f\left(-x\right)=3\left(-x\right)^2=3x^2=f\left(x\right)$f(x)=3(x)2=3x2=f(x)

Even functions possess symmetry across the $y$y axis - the part of the curve that is to the left of the $y$y axis is a mirror image of the part of the curve to the right.

## A word of caution

Oddness and evenness can be problematic in cases where the exponent $r$r is not a whole number.

We can show, for example, that $f\left(x\right)=x^{1.2}$f(x)=x1.2 is even with the following careful reasoning:

 $f\left(-x\right)$f(−x) $=$= $\left(-x\right)^{1.2}$(−x)1.2 $=$= $\left(-x\right)^{\frac{6}{5}}$(−x)65​ $=$= $\left(-1\right)^{\frac{6}{5}}\left(x\right)^{\frac{6}{5}}$(−1)65​(x)65​ $=$= $\left[\left(-1\right)^{\frac{1}{5}}\right]^6\times\left(x\right)^{\frac{6}{5}}$[(−1)15​]6×(x)65​ $=$= $\left[\sqrt{-1}\right]^6\times\left(x\right)^{\frac{6}{5}}$[5√−1]6×(x)65​ $=$= $\left[-1\right]^6\times\left(x\right)^{\frac{6}{5}}=\left(x\right)^{\frac{6}{5}}=f\left(x\right)$[−1]6×(x)65​=(x)65​=f(x)

Note in the second last line that $\sqrt{-1}=-1$51=1 and this allowed the surd to be evaluated as a real number. It would not have worked for, say, $\sqrt{-1}$61 .

Here is the correct graph of $f\left(x\right)=x^{1.2}$f(x)=x1.2 #### Worked Examples

##### QUESTION 1

If $f\left(x\right)=x^{2.11}$f(x)=x2.11, find $f\left(1.3\right)$f(1.3) correct to three decimal places.

##### QUESTION 2

If $f\left(x\right)=x^{-1.75}$f(x)=x1.75, find $f\left(3.8\right)$f(3.8) correct to three decimal places.

##### QUESTION 3

Consider the function $g\left(x\right)=ax^3-3x+5$g(x)=ax33x+5.

1. Determine $g\left(k\right)$g(k).

2. Form an expression for $g\left(-k\right)$g(k).

3. Is $g\left(k\right)=g\left(-k\right)$g(k)=g(k)?

Yes

A

No

B
4. Is $g\left(k\right)=-g\left(-k\right)$g(k)=g(k)?

Yes

A

No

B