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Power Functions: Definition and Differentiation, Slides of Algebra

How to define and differentiate power functions for any real number a, using the relationship between ex and ln x. It covers the cases when x is a positive real number, zero, and a negative number, as well as examples and formulas for derivatives.

What you will learn

  • How is the power function xa defined for negative numbers x?
  • What is the relationship between ex and ln x that allows us to define power functions?
  • How is the power function xa defined for positive real numbers x?

Typology: Slides

2021/2022

Uploaded on 09/27/2022

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General Power Functions
Defining General Power Functions
In previous lectures, we discussed the properties and derivatives of positive power functions and negative
power functions. Today, we discuss power functions in general.
A power function is a function of the form
f(x) = xa,
where ais any real number. We understand intuitively what it means to raise xto the power of a natural
number n: we just multiply ncopies of xtogether. We know what it means to raise xto the npower:
just divide 1 by xn. Now we are asked to raise xto the power of a, where ais any real number, but what
does, say, 3πmean? It does not make sense to multiply πcopies of 3 together. We have been ignoring this
question since we defined exponential functions, but now we intend to give you an answer.
First, restrict xto the positive real numbers and consider the exponential function exand the natural
logarithmic function ln x. There is a great relationship built into exand ln x: if we compose these two
functions with each other in either order, the resulting function is x:
eln x=xand ln(ex) = x.
This relationship is built into the definition of the natural logarithmic function: the quantity ln xis defined
to be the number such that e raised to the power of that number is x. Thus we get the first equation above
automatically. The second equation comes from asking the question: to what number must we raise e to get
ex? Obviously, the answer to this question is x, and thus the natural logarithm of exis x. We say the the
functions exand lnxare inverse functions of each other: if we take xand first apply ex, then ln x, or the
other way around, then the result is the same as if we did nothing to xat all. In a sense, exand ln xcancel
each other out.
Still considering only positive real number x, consider the power function xa, where ais any real number.
We can use the fact that exand ln xare inverses of each other to rewrite the power function xa:
xa=¡eln x¢a= ealn x.
This last expression, ealn xis what we will use to define xafor positive real numbers x. That is, we define
the quantity xafor positive real numbers xand all real numbers ato be e raised to the power of the quantity
atimes the natural logarithm of x. In this way, we know how to compute, say, 3π, because we know what
it means to take the natural logarithm of 3, multiply it by π, and then raise e to the power of that number.
Thus, to the nearest thousandth, we have that
3π= eπln 3 = e3.142·1.099 = e3.451 = 31.544.
You may be thinking at this point that we merely shifted the problem from understanding how to raise
xto the power of ato understanding how to raise e to some power. How do we know how to calculate ea,
where ais any real number? There is an answer to this question, which is somewhat beyond the scope of
this course, but here it is anyway: there is another way to write ea. Specifically, we can write eaas
ea= 1 + a+a2
2+a3
6+a4
24 +···+an
n!+··· .
We will not tell you why eais equal to this infinite sum, this series as mathematicians would say, but it is
true, and you can verify it using any calculator. Specifically, you can show that
e = e1= 1 + 1 + 1
2+1
6+1
24 +· ·· +1
n!+··· ,
and, in fact, this is where the mysterious number e comes from. You do need to know all of this for this
course, but it does justify why we know how to raise e to any power a: we can compute the infinite sum
above (or, at least, approximate it) and get the same number.
So, at this point, we know how to define xawhen xis a positive real number. What about when xequals
0? To define 0a, we take the right limit of xaas xapproaches 0. There are three cases:
1
pf3
pf4

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General Power Functions

Defining General Power Functions

In previous lectures, we discussed the properties and derivatives of positive power functions and negative power functions. Today, we discuss power functions in general. A power function is a function of the form

f (x) = xa,

where a is any real number. We understand intuitively what it means to raise x to the power of a natural number n: we just multiply n copies of x together. We know what it means to raise x to the −n power: just divide 1 by xn. Now we are asked to raise x to the power of a, where a is any real number, but what does, say, 3π^ mean? It does not make sense to multiply π copies of 3 together. We have been ignoring this question since we defined exponential functions, but now we intend to give you an answer. First, restrict x to the positive real numbers and consider the exponential function ex^ and the natural logarithmic function ln x. There is a great relationship built into ex^ and ln x: if we compose these two functions with each other in either order, the resulting function is x:

eln^ x^ = x and ln(ex) = x.

This relationship is built into the definition of the natural logarithmic function: the quantity ln x is defined to be the number such that e raised to the power of that number is x. Thus we get the first equation above automatically. The second equation comes from asking the question: to what number must we raise e to get ex? Obviously, the answer to this question is x, and thus the natural logarithm of ex^ is x. We say the the functions ex^ and ln x are inverse functions of each other: if we take x and first apply ex, then ln x, or the other way around, then the result is the same as if we did nothing to x at all. In a sense, ex^ and ln x cancel each other out. Still considering only positive real number x, consider the power function xa, where a is any real number. We can use the fact that ex^ and ln x are inverses of each other to rewrite the power function xa:

xa^ =

eln^ x

)a = ea^ ln^ x.

This last expression, ea^ ln^ x^ is what we will use to define xa^ for positive real numbers x. That is, we define the quantity xa^ for positive real numbers x and all real numbers a to be e raised to the power of the quantity a times the natural logarithm of x. In this way, we know how to compute, say, 3π^ , because we know what it means to take the natural logarithm of 3, multiply it by π, and then raise e to the power of that number. Thus, to the nearest thousandth, we have that

3 π^ = eπ^ ln 3^ = e^3.^142 ·^1.^099 = e^3.^451 = 31. 544.

You may be thinking at this point that we merely shifted the problem from understanding how to raise x to the power of a to understanding how to raise e to some power. How do we know how to calculate ea, where a is any real number? There is an answer to this question, which is somewhat beyond the scope of this course, but here it is anyway: there is another way to write ea. Specifically, we can write ea^ as

ea^ = 1 + a +

a^2 2

a^3 6

a^4 24

an n!

We will not tell you why ea^ is equal to this infinite sum, this series as mathematicians would say, but it is true, and you can verify it using any calculator. Specifically, you can show that

e = e^1 = 1 + 1 +

n!

and, in fact, this is where the mysterious number e comes from. You do need to know all of this for this course, but it does justify why we know how to raise e to any power a: we can compute the infinite sum above (or, at least, approximate it) and get the same number. So, at this point, we know how to define xa^ when x is a positive real number. What about when x equals 0? To define 0a, we take the right limit of xa^ as x approaches 0. There are three cases:

  • If a > 0, then a ln x approaches negative infinity as x approaches 0 from the right. Therefore

lim x→ 0 +^

xa^ = lim x→ 0 +^

ea^ ln^ x^ = 0,

since ex^ has a left horizontal asymptote at y = 0. Thus we say that 0a^ = 0 when a > 0.

  • If a = 0, then x^0 equals 1 for all positive values of x. Therefore the limit as x approaches 0 of x^0 is 1, so we define 0^0 = 1.
  • If a < 0, then a ln x approaches positive infinity as x approaches 0 from the right. Therefore

lim x→ 0 +^

xa^ = lim x→ 0 +^

ea^ ln^ x^ = +∞,

which means that the limit, for our purposes, is undefined. Therefore 0a^ is undefined for a < 0.

So, to summarize:

0 a^ =

0 a > 0 1 a = 0 undefined a < 0

Now for the difficult part: how do we define xa^ when x is a negative number? The answer depends very strongly on the value of a, for reasons well beyond the scope of this class having to do with complex numbers. We state the definitions of (−x)a^ for x > 0 in terms of xa, which we already know how to define, below:

  • if a is an even integer, then (−x)a^ = xa, so xa^ is an even function (as we already know).
  • if a is an odd integer, then (−x)a^ = −(xa), making xa^ an odd function.
  • if a is a rational number, and, when we write a as a reduced fraction, the denominator of a is even, then (−x)a^ is undefined.
  • if a is a rational number, and, when we write a as a reduced fraction, the denominator of a is odd, then there are two possibilities: - if the numerator of the reduced fraction is even, then (−x)a^ = xa. - if the numerator of the reduced fraction is odd, then (−x)a^ = −(xa).
  • if a is an irrational number, then xa^ is undefined.

So, for example:

  • (−2)−^4 = 2−^4 = 161.
  • (−2)−^3 = −(2−^3 ) = − 18.
  • to find (−2)

(^1016) , we write 1016 as a reduced fraction:

10 16

The denominator of 58 is even, so (−2) (^1016) is undefined.

  • to find (−2) 188 , we write 188 as a reduced fraction:

8 18

The denominator of 49 is odd, and its numerator is even, so (−2) 188 =

, whatever that number might be.

As for the domain of h(x), we note that

x is only defined for x ≥ 0, but 1 + x^2 is always greater than or equal to 1, so h(x) is defined everywhere. It also turns out that h(x) has a derivative everywhere, but this is harder to show. Finally, take k(x) = 6

cos x − 3, which we can also write as (cos x − 3) (^16)

. Sometimes it is better to find the domain of a function first, and this is one of those occasions. The outside function, x (^16) , is defined for x ≥ 0, so in order for x to be in the domain of k(x), we must have that cos x − 3 ≥ 0. We can rewrite this as cos x ≥ 3, which is impossible, because the maximum value of cos x is 1. Therefore k(x) is not defined anywhere, and so it is not differentiable anywhere either. This example illustrates the idea that being able to write down a formula does not automatically imply that that formula defines a function with a non-empty domain.