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# S2A - SOLUTIONS TO EXAM 2A 1[16 Marks Use the deﬁnition...

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Unformatted text preview: SOLUTIONS TO EXAM 2A 1. [16 Marks] Use the deﬁnition of the derivative (ﬁrst principles) to ﬁnd the derivative of f (LB) 2 V 1 — 36?. YOU MAY NOT USE DERIVATIVE FORMULAS TO ANSWER THIS QUESTION. Solution By the deﬁnition of the derivaitve: 1H0 h 1H0 h _ 1im(/1—(\$+h)2—\/1—5U2 . (/1—(1n+h)2+\/1—\$2 hHO h Wit m : lim 1 (a: I h)2 (1 \$2) : lim 1 (m2 i 21% i [22) 1 i x2 h—>0h(\/W+W) 1H0 h( 1—(\$+h)2r~m) , 722011 i h? , ih(2m + h) — hm — hm h—*0 h( 1 — (at + h)2 7* W) h—*0 MW *a W) : lim 251? l h : 2:17 we W m W+ W 2a: :2 2x/1— x2 \/1— xi? 2. [13 Marks] Determine Whether the function 9(33) 2 tan {7; has a cusp, vertical cusp or vertical tangent at x = 0. J USTIFY YOUR ANSWER. Solution By the chain rule: g’(ac) : sec2 ﬁg ((3/5) 2 sec2 ﬁg (321/3) 2 sec2 \B/E%J:’2/3 2 2232/3 sec2 6/; Since g’(\$) is continuous for a: 75 O7 .q’_(0) = \$113514qu = (+00)(1) = "00, .q’+(0) = 0613; 9’06) = (+00)(1) = "00. Therefore, g has a vertical tangent at .7: = 0. 3. [16 Marks] Find the equation of the tangent line to the curve x2 + y2 2 any + 7 at the point (3, 2). Solution Use implicit differentiation. d i d E (\$2 + 3/2) I E (553/ + 7) dy dy 2, 2 — : — CI, + ydm y + xdcc dy dy dy 2 _ z ——2 — z — 12 :5 y \$da: ydm d:c(\$ y) @ _ 23: — y (1.2: _ a: — 2y At (3,2) we have a: : 3 and y : 2. Hence :4; : E : —4 at (3,2). By the point—slope formula the tangent line to this curve at (37 2) has equation y=274(\$73)=l474\$. 4. [7 Marks] State the Mean Value Theorem. Solution Assume that f is a continuous function with domain [a7 b]. Assume that f is differentiable on (a7 b). Then there is at least one number 0 E ((1,1)) such that f (b) — f (a) / 2 — f (C) b _ a ' 5. [14 Marks] Find and identify all the critical points of f(a:) : (9:3 — 1]2/3. Solution By the chain rule: 2 d 2 2x2 I _ 3 —1/3 3 _ 2 _ f (x) — 3(1” 1) dx (x 1) — 3(503 i 1)1/3 <3\$ )— (x3 n1)1/3 Thus f has two critical points: at :C = 0 where f’(\$) is zero and at a: : 1 where f’(a:) does not exist. Observe that f’(:1:) has the same sign as :C — 17 i.e. positive for a: > 1 and negative for a: < 1. See the diagram below. Hence the signe of f’ (as) does not change at a: = 0, and m = 0 is not a local extremum by the First Derivative Test. At \$ = 1 the sign of f’(a:) changes from negative to positive. Hence f has a local minimum at a: = 1 by the First Derivative Test. -------- O--~-----~DNE+++++ Signf'(x) . x decreasing 0 decreasing 1 increasing local minimum 6. [14 Marks] Find all the inﬂection points of g(\$) = m5 — 5:104 + 3:10 — 7. JUSTIFY YOUR ANSWER. Solution g’(a:) : 53:4 — 203:3 + 3. Hence 9%) = 20273 7 602:2 = 20199: 7 3). Note g”(:r) is negative for a: < 3 and positive for a: > 3. See the diagram below. Hence :1: : 3 is an inflection point of 9 because the concavity changes from concave down to concave up at :0 = 3. Note that g”(0) = 0. Nevertheless a: = 0 is not an inflection point of 9 because the concavity does not change there: 9 is concave down for :1: < O and convave down for 0 < :1: < 3. """ 0"""""O++++++++ signgI/(X) . x concave O concave 3 concave down down inﬂection up pomt , . 3:2 — 16 7. [20 Marks] Con51der the function f (x) : 2 4 . 1T _ (a) Find all the x—intercepts of f. Solution 0 : f(0) when x2 — 16 : 07 i.e. the :ciintercepts are a: : —4 and a: = 4. (b) Find the y—intercept of f. Solution The yiintercept is f(0) : 77—346 : 4. (c) Find all the vertical asymptotes of the graph of f. Solution The veritcal asymptotes occur when the denominator of f (:13) is zero, i.e. when x2 i 4 = 0. Thus the graph of f has vertical asymptotes ast .2: = —2 and at :1: = 2. (d) Find the horizontal asymptotes of the graph of f. Solution , :1:2 — 16 hm : wﬂoo 3:2 —4 1. Hence the graph of f has the horizontal asymptote a: = 1 on the right. 1. \$2—16_ mangloo \$2 —4 — 1. Hence the graph of 1” also has the horizontal asymptote CC : 1 on the left. (e) Determine where f is increasing and Where f is decreasing. Solution By the quotient rule: , _ (MW e 4) _ (x2 — 16)(2x) _ 24:13 f (56) (x2 _ 4>2 (x2 _ 4>2 Note f’(:1:) has the same sign as its numerator 24\$, i.e it is negative for a: < 0 and positive for m > 0. See the diagram below. Therefore7 f is decreasing for a: < 0 and increasing for a: > O. ................ ~0++++++ +++++signfl(x) ,x decreasmg '2 decreasmgO increasmg 2 increasmg local minimum (f) Find and indentify all the critical points of f. Solution f has one critical point at a: : 0 Where f’(3:) : 0. By the First Deriative Test7 f has a local minimum at a: = 0. (g) Sketch the graph of f. Solution ...
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S2A - SOLUTIONS TO EXAM 2A 1[16 Marks Use the deﬁnition...

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