)(3) using the table below.

Transcription

1 Evaluate ( g f )(). Given f ( ) = + and g( ) =. If f ( ) = + and g( ) =, then ( g f )( ) = + and ( g f )() = = = + f() Evaluate ( f g )() using the table below. g() 5 f() g() 5 ( f g)() = f( g()) = f() =. Find the function g( ) such that ( f g)( ) = ( ) if f( ) =. ( f g)( ) = ( ) if f ( ) = and g( ) =.

2 Find the domain of ( g f )( ), if f and g ( ) = ( ) =. f( ) = D = (,] R = [, ) g ( ) = D = (, ) g f g f = = = ( )( ) ( ) ( ) Since the domainof ( g f )( ) is all in the domain of f such that f ( ) is in the domain of g, then the domain of ( g f )( ) is (,]. f International Shoe Sizes: Ever notice that your athletic shoes are labeled with three different sizes? For eample mine say: US 8.5 UK and EUR. The function that relates the US sizes to the European (EUR) is g() = +, while the function that relates EURopean sizes to sizes in the United Kingdom (UK) is f() =.5-. US EUR g() = + EUR UK f() =.5- US UK h() = f(g()) = f(+)=.5(+)- =-.5 h()=-.5=.5 Find a function that relates the US size directly to the UK size. Use this function to find the UK size for a shoe that is a US. True or False: ) If an even function is shifted vertically units up, it remains an even function. ) True, a vertical shift (up or down) of an even function will still be symmetric to the y-ais. ) False, a vertical shift of an odd function will no longer be symmetric to the origin. ) If an odd function is shifted vertically units up, it remains an odd function.

3 Find the domain and range of the function which results when the following transformations are applied to f ( ) =.. f() is epanded vertically by a factor of. f() is reflected over the -ais. f() is shifted up units. f() is shifted right 5 units The resulting equation is f ( ) = 5+. This function has a domain of { 5} and a range { y y }. Consider the graph of f shown below. Use the graph to sketch the graph of a) f(+) b) f()+ a) f(+) b) f( ) + If f( ) = and g( ) =, find the intersection of - f( )+ and g( - ). - f( )+ = - + and g( - ) = Using the intersection option on the calculator, these functions intersect at (.7,.)

4 Write an equation for the graph of the function. f() The basic function is f ( ) =. This function has been a) reflected over the ais f ( ) = b) compressed vertically by a factor of / f ( ) = c) shifted up f ( ) = + d) shifted right f ( ) = NOTE: You must do reflections before you do any shifting. A rocket is launched vertically upward from the surface of Mars. The table below gives the height of the object at the indicated time following launch. Time (seconds) Height (feet) H (.8) H (.) Averagevelocity = = = =.75 feet sec Use the data to compute the average velocity of the rocket on the time interval [.,.8]. Given: f() = - Find the slope of the secant line from the point P(,) to the point Q(a,f(a)) for the following values of a: i) a= ii) a=. iii) a=. Use these results to approimate the slope of the tangent line at the point P(,). f() f() 7 i) msec = = = f(.) f(). ii) msec = = =... f(.) f(). iii) msec = = =... It appears that the slopeof the secant is iting to. Thus, the slope of the tangent line appears to be.

5 A rocket is launched vertically upward from the surface of Mars. The table below gives the height of the object at the indicated time following launch. Time (seconds) Height (feet) H () H () Averagevelocity = = = = 5. feet sec Use the data to compute the average velocity of the rocket during the first seconds of its flight. The figure shows the distance (in meters) traveled during the first 5 seconds for a Ford Mustang accelerating from a stand still. s(t) 5 s(t) 5 t 5 5 t 5 5 Estimate the Mustang s velocity after seconds have elapsed. Inserting an approimate tangent line at t = and estimating the slope of that tangent line to be: Δs meters Velocity = = 8 Δt 5 5 sec f() f() a c b Order the following in increasing order: a) m tan at = b) m sec on the interval [-,5] c) m tan at = a) m tan at = is positive b) m sec on the interval [-,5]= c) m tan at = is negative Thus, the increasing order is c,b,a 5

6 Evaluate theit, if it eists. + + ( + )( + ) ( + ) + = = = = ( + )( ) ( ) + + Evaluate + + = + Since = and =, = DNE. + Thus, + also Does Not Eist. Evaluate ( + ) = = + + ( )( + + ) ( ) ( + ) ( + ) = = = ( )( + + ) ( ) ( + + ) = = = + +

7 + Given Find the it, L. Then find d> such that f()-l <. whenever < - < d. + = + = + = Thus + <.. < + <..< + <.+.9 < <..9() < <.().8 < <. δ =.8 =. δ =. =. δ = min( δ, δ ) =.. Evaluate theit, if it eists. ( ) ( ) As ( ) < and >, thus ( ) < The numerator is always negative and so ( ) = Using the ε,δ definition of a it, determine how close must be to in order to get f()= within of 8. f ( ) L < ε whenever < a < δ becomes 8 whenever 8 7< < 9 7 < < 9.9 < <.8 δ =.9 =.87 δ =.8 =.8 < < < δ < < δ = min( δ, δ ) =.8 ( or smaller) Thus, must be within.8of =. 7

8 If g ( ) cos for all, find g( ). Since g( ) cos and and = = cos = cos = () = Then by the Sandwich (Squeeze) Theorem g ( ) =. Find all values for which 5 f( ) = is continuous. 5 f ( ) = is continuous on its domain. Needed :5 and 5 when 5 or 5 Thus, f ( ) is continuouson, (, ). 5 Use the Intermediate Value Theorem to show that = cos has a solution in the interval [,]. f ( ) = cos is the difference of two functions which are both continuous on [,]. This means f ( ) is continuous on [,] also. ** VERY IMPORTANT!!** f () = cos = = f () = cos. Since f () < < f () and f ( ) is continuous, there is a root in (,). 8

9 At = 8 the above function is continuous At = 8 the above function is NOT continuous, but is the function from the right, because f (8) = f( ). + continuous from the right, continuous from the left or neither? 8 9