Transcript n must be even

Problem of the Day Calculator
x 0 1 2
f(x) 1 k 2
The function f is continuous on the closed
interval [0, 2] and has values that are given in
the table above. The equation f(x) = ½ must
have at least two solutions in the interval [0, 2] if
k=?
A) 0
B) ½
C) 1
D) 2
E)
3
Problem of the Day Calculator
x 0 1 2
f(x) 1 k 2
The function f is continuous on the closed
interval [0, 2] and has values that are given in
the table above. The equation f(x) = ½ must
have at least two solutions in the interval [0, 2] if
B) ½
C) 1
D) 2
E)
k = ?A) 0
When3x = 1 to get ½ twice as a y value the graph must go
below
y = ½. The only value that does this is if y = 0 when x = 1.
Thomas Simpson, an English mathematician
developed a method for approximating the area
under the curve by second-degree polynomials.
(The proof is on page 311)
Simpson's rule can be
derived by
approximating ∫ f(x) (in
blue) by the quadratic ∫
P(x) (in red)
Simpson's Rule
n must be even
Example
n=8
Simpson's Rule
≈2.0003
Remember that to use Simpson's Rule, the
number of intervals must be even!
Trapezoidal vs. Simpson
If the integrand is a linear function, the trapezoidal
gives a more exact result because it is based on a
piece-wise linear function.
If the integrand is a quadratic, cubic, or "wellbehaved" function, Simpson's gives a more exact
result becaues it is based on a piece-wise
quadratic function.
absolute value of the y-coordinate of the maximum
4
point on the interval for f '' and f
Remember to find the max on the interval you
must check the endpoints too!
Example
error in
Use the trapezoidal rule to find the
2
if n = 4
To find max of f ''(x) on the interval take derivative of f '' and set equal to
0
critical number is 0
To find absolute value of max on interval, test
critical numbers and endpoints
|f ''(0)| = 1
|f ''(1)| =
max
Example
error in
Use the Simpson's rule to find the
2
if n = 4
critical number
at .68325
4
f (0) = -3
f 4(.68325) = .680725
4
f (1) = .79549
< .000065
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