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Unformatted text preview: ial of the for m a 0 + a 1 t + a 2 t 2
2
1
with coefficients a 0 = 0 , a 1 = u . s i n θ a nd a2 =   g . So the height
2
function y(t) is single valued and continuous .
single
continuous
It is important to know that polynomials are to functions what rational numbers are
to real numbers. We can approximate any real number to any required degree of
accuracy by a suitable rational number. Likewise we can approximate almost any real
value function by a suitable polynomial. For this we have Lagrange’s method of
interpolation. 42 Example 1: is f(x) continuous at x = a ?
1
continuous
f(x)
a f (x ) = { x for 0 ≤ x ≤ a
for
2a  x for a ≤ x ≤ 2 a a
(0,0
0,0)
(0,0 )
Approaching a from the left : f(x) = x
left
Near a and to the left of a :
left 2a x x = a − δx L imit f(x) = Limit { a − δx} = a
δx → 0
x → a−
Approaching a from the right : f(x) = 2a − x
right
Near a and to the right of a : x = a + δx
right L imit f(x) = Limit { 2a −(a +δx)} = Limit { 2a −a −δx }
δx → 0
δx → 0
x → a+
Limit
= δx → 0 { a − δx } = a
Hence, L imita− f(x) = a = L imita+ f(x)
x→
x→ Value f(x) = f(a) = a
x=a
Since, L imita− f(x) = L imita+ f(x) = Value f(x) = a
x→
x→
x=a
we say : f(x) is continuous at x = a.
continuous 43 Example 2 : is f(x) = x continuous at x = 0 ?
continuous
f(x) f(x) = x (0 , 0 )
Approaching 0 from the left : f(x) = x = −x
left
Near 0 and to the left of 0 :
left L imit x → 0− x x = 0 − δx Limit
Limit
f(x) = δx → 0 { − (0 − δx)} = δx → 0 { +δx} = 0 Approaching 0 from the right : f(x) = x
right
Near 0 and to the right of 0 : x = 0 + δx
right L imit x → 0+ Limit
Limit
f(x) = δx → 0 { (0 +δx) = δx → 0 { + δx } = 0 Hence, L imit − f(x) = 0 = L imit + f(x)
x→0 x→0 Value
x = 0 f(x) = f(0) = 0 Value
Since, L imit0− f(x) = L imit + f(x) = x = 0 f(x) = 0
x→
x→0
we say : f(x) is continuous at x = 0.
continuous 44 [meter s / sec] 3:
Example 3: v(t) is the speed of a car weighing one ton that star ts fro...
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 Fall '09
 TAMERDOğAN

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