Exponential Functions



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Eponential Functions Deinition: An Eponential Function is an unction that has the orm ( a, where a > 0. The number a is called the base. Eample:Let For eample (0, (, ( It is clear what the unction means or some values o. ( 4, ( 8, (, (.44,. 5 5 and (. 8. (, 4 Deining ( or irrational is too diicult now. This graph represents eponential growth since it is increasing as increases. 9. Lecture 5 o 8 =

Recall that replacing with in a unction amounts to a rotation o the graph about the -ais. So we can get the graph o previousl. It is shown dotted below. The unction ( represents eponential deca since the unction decreases as increases. (, rom = (/ = - ( shown 9. Lecture 5 o 8 =

Eponential Growth: All unctions ( a where a >, ehibit eponential growth. 4 0. 5. 9. Lecture 5 o 8

Eponential Deca: All unctions ( a where 0 < a <, ehibit eponential deca. 0. 6 0.5 0. 4 0. 0 0. 7 0. 8 0. 9 9. Lecture 5 4 o 8

9. Lecture 5 5 o 8 Alternate representation or eponential deca unctions. We can write b a a a (, where a b So that a ( where 0 < a <, can be written as b ( where b >. Eamples: 0 0 5 5.4 0

THE Eponential Function e There is onl one eponential unction that has a slope o at the point (0,. This is called the eponential unction and we denote the base with the letter e. We will ind out that e =.78888459045 0 9 ( e 8 7 6 5 4-5 -4 - - - 0 4 5 9. Lecture 5 6 o 8

9. Lecture 5 7 o 8-5 -4 - - - 0 4 5 4 5 6 7 8 9 0 Important related unctions are k e ( or real k. I k > 0, k e ( eponential growth unctions. I k < 0, k e ( eponential deca unctions. e ( e 0.5* ( e ( e 0.5* (

Inverse Functions Deinition: awe sa ( and g( are inverse to each other i g( g (, domain o = range o g 4 domain o g = range o. bthe inverse o a unction is denoted. Domain o Range o g Range o g Domain o g 9. Lecture 5 8 o 8

Eamples o Inverse Functions Tabular unctions: Just interchange the columns. ( ( 4 6 8 9 6 8 5 6 8 5 4 6 8 6 9 8 4 5 6 5 8 9. Lecture 5 9 o 8

Finding the graph o an inverse unction rom the graph o the unction. I the point (a, b is on the graph o (, then the point (b, a is on the graph o (. This represents a rotation o the graph o ( about the line =. b (b, a = a a (a, b b 9. Lecture 5 0 o 8

Eample: The unction ( has the propert that it triples the input, so it shouldn t surprise ou that the inverse unction is one that would take one third o the input, or (. This is shown in the graph below. 5 4 = -5-4 - - - 0 4 5 - - - -4-5 9. Lecture 5 o 8

Eample: Consider the unction (. Thinking about the operations ou irst double then input then add one to the result. To get the inverse unction undo the unction, i.e. irst subtract one then halve the result, or (. This is shown in the graph below. 5 4-5 -4 - - - 0 4 5 - - - -4-5 9. Lecture 5 o 8 =

When does a unction have an Inverse? For a unction to have in inverse it must pass the horizontal line test. That is, an horizontal line must cross the graph o the unction at most once. I the two dierent points ( a, b and ( a, b are both on the graph o (, then the two points ( b, a and ( b, a would be on the graph o (. But this would mean that ( would have two dierent values in its range associated with the same value in its domain. This violates the deinition o a unction. This happens with the unction (. 9. Lecture 5 o 8

5 4-5 -4 - - - 0 4 5 - - - -4-5 9. Lecture 5 4 o 8

5 4-5 -4 - - - 0 4 5 - - - -4-5 9. Lecture 5 5 o 8

Finding Inverse Functions given Epressions When given an epression (, one ma ind an inverse unction b interchanging and, then solving the resulting equation or. Eample: Find the inverse o (. So, First interchange and. This gives. Now solve or. or so (. 9. Lecture 5 6 o 8

Eample: Find the inverse o ( 4. So, 4 First interchange and. This gives 4. Now solve or. 4 4 or 4 4 so ( 4. 9. Lecture 5 7 o 8

When can this procedure break down? When ou cannot solve or. 4 Eample: Let ( 4 Then 4 4, and upon interchanging variables we get, 4 4 to be solved or. This is diicult i not impossible. When an inverse doesn t eist. Eample: Find the inverse o ( 6. Write 6, and interchange variables to give 6 6 6 or Aha, ou see, that s not a unction. O course, ou knew that ( does not have an inverse, because it lunks the horizontal line test. 9. Lecture 5 8 o 8