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. yint (x 2 0)e => f (0) = e
=
2 >0
0int because
= esin(x)
So
2. xint: (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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Full Document
 Fall '08
 HSKEL
 Math, Derivative, Fundamental Theorem Of Calculus, Sin, lim, esin

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