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B) f 0 (x) = 3cx2 ; thus f 0 (1) = 3c and f 0 (2) = 12c. 1 (c) f 0 (x) = 10x−3 ; thus f 0 (1) = 10 and f 0 (2) = 10 8 = 44 √ √ (d) f 0 (x) = x1/3 = 3 x; thus f 0 (1) = 1 and f 0 (2) = 3 2 (e) f 0 (w) = 2w−2/3 ; thus f 0 (1) = 2 and f 0 (2) = 2 · 2−2/3 = 21/3 (f) f 0 (w) = 12 w−7/6 ; thus f 0 (1) = 1 2 and f 0 (2) = 12 (2−7/6 ) = 2−1 · 2−7/6 4. 2 1. V C = Q3 − 5Q2 + 12Q. The derivative d dQ V C = 3Q2 − 10Q + 12 is the M C function. 2. C = AC · Q = Q3 − 4Q2 + 174Q. Thus M C = dC/dQ = 3Q2 − 9Q + 174.

No, because q = v + 1, as such, is defined at every value of v, whereas the given rational function is not defined at v = 2 and v = −2. The only permissible way to rewrite is to qualify the equation q = v + 1 by the restrictions v 6= 2 and v 6= −2. 6. Yes; each function is not only continuous but also smooth. 1 1. 2. (a) dy/dx = 12x11 (b) dy/dx = 0 (c) dy/dx = 35x4 (d) dw/du = −3u−2 (e) dw/du = −2u−1/2 (f) dw/du = u−3/4 (a) 4x−5 (b) 3x−2/3 (c) 20w3 (d) 2cx (e) abub−1 (f) abu−b−1 3. (a) f 0 (x) = 18; thus f 0 (1) = 18 and f 0 (2) = 18.

2 x = = 1 + 2x + 1 2 2! (2x) + 1 3 3! (2x) +···+ 1 n n! (2x) e2p n+1 (n+1)! (2x) It can be verified that Rn → 0 as n → ∞. (c) Hence φ(x) can be expressed as an infinite series: φ(x) = 1 + 2x + 1 2 2! (2x) 3. 12 4. 07 ( or 7%) + 1 3 3! 40 (d) 1 ( or 100% ) 5. When t = 0, the two functions have the same value ( the same y intercept ). Also, y1 = Aer when t = 1, but y2 = Aer when t = −1. Generally, y1 = y2 whenever the value of t in one function is the negative of the t value in the other; hence the mirror- image relationship.

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