2 3c 5 1 p c using the mvt 4 3 determine the values of

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Unformatted text preview: s2cand differentiable everywhere. 2 3c + == 5 1 p c=± Using the MVT, 4 3 determine the values of h (3) and 2 c≤ =(10) – f(2) ≤ k? k, where ch ± p2 = f3 f f (10) f (2) = f (10) 2 (3) f (10) c) = (2) 0 f( c = ± p 5 What should we pick for 10 8 2 8 (3) 1=9 2 9 f (10) f (2) 5 8 8 f (10) f (c) = 10 f (10) 0 0 f (c) = 10 8 the f (2) f (2) f (10) values a and b? = 5 f 0 (c) 8 2 8 f (2) f (10) f (2) = 2 8= ) =) f (10) f (2) 5 f 0 (c) 8 We 5 8 8 know: So: 40 f (10) f (2) 64 =) h = 40, k = 64 2 curvesketch sin(x) ( x) = Df= {x 2eR} Curve Sketching ⇡⇡ ( ,) e 2 following curve: Rsin(0)= esin(= {x 2 R} 2 D = f (0)2 f (x) = 1D x) {x =x } R} ¤༊ Sketch the D={ 2 3⇡ ⇡ 0 = esin(x) ( , = ) {x 2 R 1 1.  Domain: f (0)D2=sin(0) = } e 2 esin(x) > 0 0 3⇡ esin(x) sin(0) 5⇡ f (0) = esin(0) = 1 2.  Intercepts: = f (0) =1 D= ( , sin(x) = e ) sin(0) {x 2 R} =1 1.  y-int (x 2 0)e => f (0) = e = 2 >0 0int because = esin(x) So 2.  x-int: (y ⇡ 0) ⇡ => 0 = sin( x) sin(x)snoxx – =3 esin(x) e in( ) 6= f (x) ( ( , f) x) = e0 = e = esin(x) > 0 2 (x ) 3.  Symmetry: 2 esinin(sin(x0 >) ⇡ 5x f ( x) =3esin( ⇡ ) = e s e x) 6= > x) Not odd or f( 0 (,) 22 6= f (x) even What about a periodic f(x) is a periodic function? 6= f (x) f (x + 2⇡ ) = esin(x+2⇡) = esin(x) function with period 2π 6= f (x) ) 6= 4. Asymptotes f (x) = esin(x) 1.  Horizontal (check at +/- ∞) => None lim ff (x) = esin (1) = not defined = esin 1 ) = not defined lim (x) x!1 x!1 sin ( 1) = {x 2 R} ( 1D ) = not defined esin lim (x) = lim 1ff (x) = e x! x! 1 = not defined 2. Vertical (check at undefined points in domain) => None because f(x) is well defined everywhere 2 2 x!1 sin lim )f (x) =...
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This document was uploaded on 03/04/2014 for the course MATH 1A03 at McMaster University.

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