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MATH 136 The Inverse Function Theorem A function y = f (x) is one–to–one if it is always the case that different x values are assigned to different y values. A continuous strictly increasing (or strictly decreasing) function is one–to–one. € For example, y = x 3 + 4 is one–to–one, but y = sin x is not one–to–one. The sine function assigns many different x values to the same y . For instance 0, π, 2π, 3π, etc. all have a sine of 0. € € € 3 y = sin x y = x +4 A one–to–one function f (x) always has an inverse denoted by f −1(x) . The inverse “undoes” the steps applied to x by the function f . For instance with f (x) = x 3 + 4 , the function f first cubes, then adds 4. To undo the process, we first subtract 4, then take the cube root. Thus f −1(x) = (x − 4)1/ 3 . € One property of f and f −1 is that Domain f = Range f −1 € and Domain f −1 = Range f . This property essentially means that coordinates are “paired up.” Consider f (x) = x 3 and f −1(x) = x1/ 3. Since f (2) = 8, then f −1(8) = 2. The pair (2, 8) is created by this inverse relationship. The point 2 is in the domain of f and in the range of f −1 . € Likewise, 8 is in range of f and in the domain of f −1 . € The Derivative of the Inverse Suppose we have a one–to–one function for which we have a specific known rule, such as f (x) = x 3 + 4 . Then f −1(x) exists. We wish to find the derivative of f −1(x) at some point b in the domain of f −1 . That is, we wish to find ( f −1)′ (b) . € € Note that b is also in the range of f and is paired up with a point a in the domain of f such that f (a) = b . The Inverse Function Theorem: We find ( f −1)′ (b) in terms of f€′(a) by € € 1 , where f (a) = b . ( f −1)′ (b) = f ′(a) € Example 1. Let f (x) = x 3 + 4 . Find ( f −1)′ (220) . Solution. The Inverse Function Theorem implies that we can compute ( f −1)′ (220) −1 without actually € finding f or its derivative. Step 1: Solve f (x) = 220: x 3 + 4 = 220 gives x 3 = 216 and x = 216 3 = 6. Thus f (6) = 220 and the pair (6, 220) is created. 2 Step 2: f ′(x) = 3x €and f ′(6) = 108. € Step 3: ( f€−1)′ € (220) = 1 1 = . f ′(6) 108 In this case, we actually can verify the answer by finding f −1 and its derivative. Here f −1(x) = (x − 4)1/ 3 . ( f −1)′ (220) = Thus, ( f −1)′ (x ) = 1 1 = . 2 / 3 108 3(220 − 4) 1 1 . (x − 4) −2/ 3 = 3 3(x − 4)2 / 3 Thus, 3x + 15 − 21 . Find ( f −1)′ (−9) using the Inverse Function 2 Theorem. Verify the result by finding f −1(x) and its derivative. Example 2. Let f (x) = 3x + 15 − 21 = –9 gives 3x + 15 = 3 , which gives 2 3x + 15 = 9 and x = –2. Thus f (−2) = –9 and the pair (–2, 9) is created. Solution. We first solve f (x) = –9: 3 1 21 1 Re-writing f (x) = 3x + 15 − , we have f ′(x) = (3x + 15)−1/ 2 × 3 = 4 3x + 15 2 2 4 € 1 1 −1 and f ′(−2) = . Thus, ( f )′ (−9) = = 4. f ′(−2) 4 To find f −1(x) , we first observe the steps used to create f (x) : multiply by 3, add 15, take square root, subtract 21, divide by 2. To create f −1(x) , we undo these steps in reverse order: Given x , multiply by 2, add 21, square, subtract 15, divide by 3. Thus, € Then ( f −1)′ (x ) = f −1(x) = (2x + 21)2 − 15 3 = 1 ( 2x + 21)2 − 5 . 3 2 4 4 ( 2x + 21) × 2 = (2x + 21) and ( f −1)′ (−9) = (−18 + 21) = 4. 3 3 3 The Inverse Trig Functions The trigonometric functions cos x , sin x , tan x are not one-to-one; thus they do not have inverses. However we can make them one-to-one by restricting the domains to only two quadrants so that we only have one segment of each graph. € €the domain € With cos x , we restrict to [0, π] (the 1st and 2nd quadrants). Then the −1 “inverse cosine” exists: cos x is the angle (in radians) whose cosine is x , where the angle must be in either the 1st or 2nd quadrant. € € € f (x) = cos x Domain: 0 ≤ x ≤ π Range: –1 ≤ y ≤ 1 x€ € 0 π 6 π 4 π 3 π 2 2π 3 3π 4 5π 6 π −1 f (x) = cos x Domain: –1 ≤ x ≤ 1 Range: 0 ≤ y ≤ π cos x 1 3 2 2 2 1 2 0 1 – 2 2 – 2 3 – 2 –1 x € € cos −1 x 1 3 2 2 2 1 2 0 π 6 π 4 π 3 π 2 2π 3 3π 4 5π 6 π 0 1 – 2 2 – 2 3 – 2 –1 With sin x , we restrict the domain to [–π/2, π/2] (the 1st and 4th quadrants). Then the “inverse sine” exists: sin−1 x is the angle (on radians) whose sine is x , where the angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then we € write it as a negative angle. € € f (x) = sin x Domain: –π/2 ≤ x ≤ π/2 Range: –1 ≤ y ≤ 1 x € π 2 π – 3 π – 4 π – 6 – 0 π 6 π 4 π 3 π 2 sin x −1 f (x) = sin x Domain: –1 ≤ x ≤ 1 Range: –π/2 ≤ y ≤ π/2 x € –1 3 – 2 2 – 2 1 – 2 –1 3 – 2 2 – 2 1 – 2 0 1 2 2 2 3 2 0 1 2 2 2 3 2 1 1 sin −1 x π 2 π – 3 π – 4 π – 6 – 0 π 6 π 4 π 3 π 2 With tan x , we restrict the domain to (–π/2, π/2) (the 1st and 4th quadrants). Then the “inverse tangent” exists: tan−1 x is the angle (in radians) whose tangent is x , where the angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then € we write it as a negative angle. € € f (x) = tan x Domain: –π/2 < x < π/2 Range: –∞ < y < ∞ € −1 f (x) = tan x Domain: –∞ < x < ∞ Range: –π/2 < y < π/2 € x π 2 π – 3 π – 4 π – 6 – 0 π 6 π 4 π 3 π 2 tan x x –∞ –∞ – 3 – 3 –1 1 – 3 –1 1 – 3 0 1 3 0 1 3 1 1 3 3 ∞ ∞ tan −1 x π 2 π – 3 π – 4 π – 6 – 0 π 6 π 4 π 3 π 2 Example 3. (a) Let f (x) = cos x for 0 ≤ x ≤ π . Find ( f −1)′ (1 / 2) . (b) Let f (x) = sin x for −π /2 ≤ x ≤ π /2 . Find ( f −1)′ (− 2 / 2) . € € We first solve cos x = 1/2, to get x = π /3. Then f ′(x) = −sin x and € € 1 −2 3 . Thus, ( f −1)′ (1 / 2) = = . f ′(π / 3) = − sin(π / 3) = − f ′(π / 3) 3 2 € € (b) We solve sin x = − 2 / 2 , to get x = −π /4 . Then f ′(x) = cosx and 1 2 2 . Thus, ( f −1)′ (− 2 / 2) = = . f ′(−π / 4) = cos(−π / 4) = f ′(−π / 4) 2 2 € € Solution. (a)