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Mathematics

Inverse Functions

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Inverse Functions Given a function , if there is a function such that which equals the identity function The function is said to be "invertible" "undoes" what "does", and vice versa Such a function is called the inverse, denoted as (the inverse of ) The notation is not to be confused with an exponent In some cases the inverse of a function can be found through algebraic methods CONSIDER: Given to determine we must find a functions that must undo But, recall the set of outputs from , to undo we take Thus Observe that: Not every function has an inverse CONSIDER: But is not a function. An inverse only exists when different inputs in the domain always yield different outputs in the range Such functions are called one-to-one

Ch9 SG

Subject: 
Calculus
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372 CHAPTER 9 Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. y? = 2x2ex 3/3 = x2y and y(0) = 2 by inspection. 2. y? = x3 ? 2 sinx, y(0) = 3 by inspection. 3. (a) ?rst order; dy dx = c; (1 + x) dy dx = (1 + x)c = y (b) second order; y? = c1 cos t? c2 sin t, y?? + y = ?c1 sin t? c2 cos t+ (c1 sin t+ c2 cos t) = 0 4. (a) ?rst order; 2 dy dx + y = 2 ( ? c 2 e?x/2 + 1 ) + ce?x/2 + x? 3 = x? 1 (b) second order; y? = c1et ? c2e?t, y?? ? y = c1et + c2e?t ? ( c1et + c2e?t ) = 0 5. 1 y dy dx = x dy dx + y, dy dx (1? xy) = y2, dy dx = y2 1? xy 6. 2x+ y2 + 2xy dy dx = 0, by inspection. 7. (a) IF: ? = e3 ? dx = e3x, d dx [ ye3x ] = 0, ye3x = C, y = Ce?3x separation of variables: dy y = ?3dx, ln |y| = ?3x+ C1, y = ?e?3xeC1 = Ce?3x

Ch8 SG

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317 CHAPTER 8 Principles of Integral Evaluation EXERCISE SET 8.1 1. u = 3? 2x, du = ?2dx, ? 1 2 ? u3 du = ?1 8 u4 + C = ?1 8 (3? 2x)4 + C 2. u = 4 + 9x, du = 9dx, 1 9 ? u1/2 du = 2 3 ? 9u 3/2 + C = 2 27 (4 + 9x)3/2 + C 3. u = x2, du = 2xdx, 1 2 ? sec2 u du = 1 2 tanu+ C = 1 2 tan(x2) + C 4. u = x2, du = 2xdx, 2 ? tanu du = ?2 ln | cosu |+ C = ?2 ln | cos(x2)|+ C 5. u = 2 + cos 3x, du = ?3 sin 3xdx, ? 1 3 ? du u = ?1 3 ln |u|+ C = ?1 3 ln(2 + cos 3x) + C 6. u = 3x 2 , du = 3 2 dx, 2 3 ? du 4 + 4u2 = 1 6 ? du 1 + u2 = 1 6 tan?1 u+ C = 1 6 tan?1(3x/2) + C 7. u = ex, du = exdx, ? sinhu du = coshu+ C = cosh ex + C 8. u = lnx, du = 1 x dx, ? secu tanu du = secu+ C = sec(lnx) + C 9. u = cotx, du = ? csc2 xdx, ? ? eu du = ?eu + C = ?ecot x + C

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