Lesson 15: Inverse Functions and Logarithms

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The inverse of a function "undoes" the effect of the function. We look at the implications of that property in the derivative, as well as logarithmic functions, which are inverses of exponential functions.

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Lesson 15: Inverse Functions and Logarithms

  1. 1. Section 3.2 Inverse Functions and Logarithms V63.0121.034, Calculus I October 21, 2009 Announcements Midterm course evaluations at the end of class . . Image credit: Roger Smith . . . . . .
  2. 2. Outline Inverse Functions Derivatives of Inverse Functions Logarithmic Functions . . . . . .
  3. 3. What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. . . . . . .
  4. 4. What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. So f−1 (f(x)) = x, f(f−1 (x)) = x . . . . . .
  5. 5. What functions are invertible? In order for f−1 to be a function, there must be only one a in D corresponding to each b in E. Such a function is called one-to-one The graph of such a function passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is continuous, then f−1 is continuous. . . . . . .
  6. 6. Graphing an inverse function The graph of f−1 interchanges the x and y f . coordinate of every point on the graph of f . . . . . . .
  7. 7. Graphing an inverse function The graph of f−1 interchanges the x and y f . coordinate of every point on the graph of f .−1 f The result is that to get the graph of f−1 , we . need only reflect the graph of f in the diagonal line y = x. . . . . . .
  8. 8. How to find the inverse function 1. Write y = f(x) 2. Solve for x in terms of y 3. To express f−1 as a function of x, interchange x and y . . . . . .
  9. 9. How to find the inverse function 1. Write y = f(x) 2. Solve for x in terms of y 3. To express f−1 as a function of x, interchange x and y Example Find the inverse function of f(x) = x3 + 1. . . . . . .
  10. 10. How to find the inverse function 1. Write y = f(x) 2. Solve for x in terms of y 3. To express f−1 as a function of x, interchange x and y Example Find the inverse function of f(x) = x3 + 1. Answer √ y = x3 + 1 =⇒ x = 3 y − 1, so √ f−1 (x) = 3 x−1 . . . . . .
  11. 11. Outline Inverse Functions Derivatives of Inverse Functions Logarithmic Functions . . . . . .
  12. 12. derivative of square root √ dy Recall that if y = x, we can find by implicit differentiation: dx √ y= x =⇒ y2 = x dy =⇒ 2y =1 dx dy 1 1 =⇒ = = √ dx 2y 2 x d 2 Notice 2y = y , and y is the inverse of the squaring function. dy . . . . . .
  13. 13. Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) . . . . . .
  14. 14. Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) “Proof”. If y = f−1 (x), then f (y ) = x , So by implicit differentiation dy dy 1 1 f′ (y) = 1 =⇒ = ′ = ′ −1 dx dx f (y) f (f (x)) . . . . . .
  15. 15. Outline Inverse Functions Derivatives of Inverse Functions Logarithmic Functions . . . . . .
  16. 16. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . . . . . . .
  17. 17. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . Facts (i) loga (x · x′ ) = loga x + loga x′ . . . . . .
  18. 18. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . Facts (i) loga (x · x′ ) = loga x + loga x′ (x) (ii) loga ′ = loga x − loga x′ x . . . . . .
  19. 19. Logarithms Definition The base a logarithm loga x is the inverse of the function ax y = loga x ⇐⇒ x = ay The natural logarithm ln x is the inverse of ex . So y = ln x ⇐⇒ x = ey . Facts (i) loga (x · x′ ) = loga x + loga x′ (x) (ii) loga ′ = loga x − loga x′ x (iii) loga (xr ) = r loga x . . . . . .
  20. 20. Logarithms convert products to sums Suppose y = loga x and y′ = loga x′ ′ Then x = ay and x′ = ay ′ ′ So xx′ = ay ay = ay+y Therefore loga (xx′ ) = y + y′ = loga x + loga x′ . . . . . .
  21. 21. Example Write as a single logarithm: 2 ln 4 − ln 3. . . . . . .
  22. 22. Example Write as a single logarithm: 2 ln 4 − ln 3. Solution 42 2 ln 4 − ln 3 = ln 42 − ln 3 = ln 3 ln 42 not ! ln 3 . . . . . .
  23. 23. Example Write as a single logarithm: 2 ln 4 − ln 3. Solution 42 2 ln 4 − ln 3 = ln 42 − ln 3 = ln 3 ln 42 not ! ln 3 Example 3 Write as a single logarithm: ln + 4 ln 2 4 . . . . . .
  24. 24. Example Write as a single logarithm: 2 ln 4 − ln 3. Solution 42 2 ln 4 − ln 3 = ln 42 − ln 3 = ln 3 ln 42 not ! ln 3 Example 3 Write as a single logarithm: ln + 4 ln 2 4 Answer ln 12 . . . . . .
  25. 25. “ . . lawn” . . Image credit: Selva . . . . . .
  26. 26. Graphs of logarithmic functions y . . = 2x y y . = log2 x . . 0 , 1) ( ..1, 0) . ( x . . . . . . .
  27. 27. Graphs of logarithmic functions y . . = 3x= 2x y . y y . = log2 x y . = log3 x . . 0 , 1) ( ..1, 0) . ( x . . . . . . .
  28. 28. Graphs of logarithmic functions y . . = .10x 3x= 2x y y=. y y . = log2 x y . = log3 x . . 0 , 1) ( y . = log10 x ..1, 0) . ( x . . . . . . .
  29. 29. Graphs of logarithmic functions y . . = .10=3xx 2x y xy y y. = .e = y . = log2 x y . = ln x y . = log3 x . . 0 , 1) ( y . = log10 x ..1, 0) . ( x . . . . . . .
  30. 30. Change of base formula for exponentials Fact If a > 0 and a ̸= 1, then ln x loga x = ln a . . . . . .
  31. 31. Change of base formula for exponentials Fact If a > 0 and a ̸= 1, then ln x loga x = ln a Proof. If y = loga x, then x = ay So ln x = ln(ay ) = y ln a Therefore ln x y = loga x = ln a . . . . . .

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