1 Theorem 1: Intermediate Value Theorem Definition: A function y=f(x) that is continuous on a closed interval [a,b] takes on every value between f(a) and f(b). If u is a number between f (a) and f (b), f (a) < u <f (b) Then there is a c∈ (a, b) such that f (c) = u. Algebraic Example:
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Theorem 1: Intermediate Value Theorem
Definition: A function y=f(x) that is continuous on a closed interval [a,b] takes on every value between f(a) and f(b).
Ifuis a number between f (a) andf (b), f (a) 0 then f(x) is concave up
f (x) is decreasing or f (x) 0(c) x < -2 or x > -2/3 (d) x2 (e)2/3 < x < 2
Correct answer is e
2.
The graph of the function f is shown above. For which of the following values of x if f (x) positive and increasing(a)a (b)b (c)c (d)d (e)ee b/c f is increasing thus f (x) is positive and f is concave up thus f (x) is increasing
Theorem 5: Extreme Value TheoremDefinition: if a real-valued function f is continuous in the closed and bounded interval [a,b], then f must attain a maximum and a minimum, each at least once. That is, there exist numbers c and d in [a,b] such that
Algebraic Example:
Find the Min and Max values of on [-2,2]
Because x=9/4 is not in the interval [2,2], the only critical point occurs at x = 0 which is (0,1). The function values at the endpoints of the interval are f(2)=9 and f(2)=39; hence, the maximum function value 39 at x = 2, and the minimum function value is 9 at x = 2. ExtremaAbsolute vs. Relative Extrema:Let f be a function defined on some interval and c be a number in that interval.
An absolute maximum of the function f occurs at (c,f(c)) if for every x value in the interval.
An absolute minimum of the function f occurs at (c,f(c)) if for every x value in the interval.
The point (c,f(c)) is a relative maximum of a function f if there exists an open interval a,b in the domain of f containing c such that for all x in a,b
The point (c,f(c)) is a relative maximum of a function f if there exists an open interval a,b in the domain of f containing c such that for all x in a,b
3 Ways to Find Extrema:Comparison of Function values:
Abs Max Value = 22 at x =3Relative Max =-.0887 at x=
Abs Min value = -20 at x=-3Relative Min=2.0887 at Abs Max b/c highest y value Relative Max b/c f (x) goes + to Abs Mix b/c lowest y value Relative Min b/c f (x) goes to +
First Derivative Test:If f (c) = 0 or DNE and f (x) changes from + to at x=c, then f(c) is a relative max.
If f(c) = 0 or DNE and f (x) changes from to + at x=c, then f(c) is a relative min.
Max b/c it goes +to - on f Min b/c it goes to +
Second Derivative Test:
0A(t)=0
V(t) < 0Speeding UpSlowing DownConstant Velocity Left
V(t) > 0Slowing DownSpeeding UpConstant Velocity Right
V(t) = 0 StoppedAccel LeftStoppedAccel RightStoppedNo Accel
Displacement vs. Total Distance:
FRQ #42005 AB 3
A particle moves along the x-axis so that its velocity v at time t, for is given by. The particle is at the position x=8 at time t=0(a) Find the acceleration of the particle at time t=4(b)
Find all times t in the open interval at which the particle changes direction. During which time intervals , for does the particle travel to the left(c) Find the position of the particle at time t=2(d) Find the average speed of the particle over the interval Solution:(a) a(4)=v (4)=(b)
The particle changes direction @ t=1,2The particle travels to the left when 1 < t 0 : (-2,2) , (2,3.5)(b) g decreasing : g0: [-7,-2] [-2,0] [2,3](f) Concave down : g decreasing , g0 : (a,c) (e, ) (above x-axis)(b) f(x) decreasing : f
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