MAT 113 Exam 1 with Solution, Fall 2007

Notre Dame University–Louaize · Intensive Calculus · 7 pages · with solution

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MAT 113 Exam 1 with Solution, Fall 2007, page 1

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x? —4x4+4 x? 45x? -14x 1-(10 points) Find the following limits: Q) 4 lL, nae C a) lim, ,. — x-2 -4 (Giver reasons for your answers) Cone c >) Coy u wafiteed = Cho vs discad hye 2. 2-(10 points) Can f(x) = be extended to be continuous at x=2 or x= -2? GS K-52 ° a> 2 Cx 2x42) XY KxD q & Roo Can \e oxtoncleL he be cud A W225 aad OF extend Poahan ke Foye fF Ka? § x42 = 5 K-22. «Gey - — gl a : \ zn ( 5 =) BEB u “Bele ined =» ¢ (x) dinate xa rf. OZ ay 8 a) Agen net emg X game 2 ‘ - WT Ce “wK i ¢ xb uA p (x) h ke Cah Cond, am 3-10 points) Show that f(x) =x? —2x +2 has a root between -2 ay iG C oy Tee wi x+. 2 . cs Cos « x3 2 M2 F cond ba lo2te ) \ (aud Colas oF ann = Gem FO =e 256 - oe +. leer he vols jee pr pee e € C © i. 0 Here bo aheet a € (- 2,0) Such Pat (0 oc) = ° 4-(15 points) Find the derivatives of the following functions: 2 = (. 9 2. --| > i (2 4: , i a) f() == 8? = + 3Xo- ax+t JX = x 3 X 2 2 47x \ Hl ; 3 =a Wie 3 . 5 (x2= UV a bY -2vw Te UX eR L 2 5-(12 points) Use the definition to find the derivative for f(x) = 2x? +5,then find the equation of the tangent line to the curve of f(x) at the point of abscissa x=2. A \ \" (Care — Rc K+ Wy oe -\ 2¢x x vey 4&5 _( 2xt £5) = - ys. \ = 9) a x teahxalh) 4 +5_ 2x26 dy | ; \ ATES PC =k ene ee ea - ZF vo ; _ a) ohn ie 4h) 7 i Vx p27 OMX. = _* ho pt J bake opt pevlact - Calis. 17) = (2,13) r= Sue of the trogiat Mp Pee) = 402) = g. eq J Foe Tang ent May 13 28 (x-2) J = 8X _\ 64\3 [o = 8x 3 | . ints) Find the equation of the normal line to the curve x? —3xy + y’ =—1 at the point (1, 1). Sa {a 4 y Dae fos ix qh , al bus Baar h Gi,1) Ne / Sym). stone of The ae! Se “A Yul =) ele oo menal coer. 7-(15 points) Find the slope for the line tangent to the curve x= 7 tant, ead at 2 4 Also, find the value of + at this point. 8-(13 points) for what values of m, if any, is f(x) = { a) Continuous at x=0? Ww, ei sin(2x), x<0 mx, x>0 b) Differentiable at x=0? [EE 5D | 7, re > a =) Qc o7) = _ ane = Sk X— > 6- DYE SOe= be— bon ag Cee) = ae C5. et)= 0 = Cees = Stn (ey