My Math Forum Average Rate of Change & Composite Functions

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 July 22nd, 2012, 10:59 AM #1 Member   Joined: Jul 2012 Posts: 50 Thanks: 0 Average Rate of Change & Composite Functions http://imageshack.us/photo/my-images/801/1abpng.jpg/ 1 http://imageshack.us/photo/my-images/696/23png.jpg/ 2 http://imageshack.us/photo/my-images/201/75785195.jpg/ 3 http://imageshack.us/photo/my-images/593/10920120.jpg/ 4 http://imageshack.us/photo/my-images/62/42820430.jpg/ 5 I've attemption most of these, others have left me confused. I want to make sure im doing all these correct exam is nearing! : ( 1ab) =[f(x2)-f(x1) ] / [ (x2)-(x1)] =[f(-1)-f(-4) ] / [ (-1)-(-4)] =[f(-1/-1-2)-f(-4/-4-2) ] / [ (3)] =[(1/3)-(2/3)] / [(2)] =[ (-1/3)/(3)] =[-1/9] therefore average rate of change on the interval -4<= x <= -1 is -1/9 b)You can't because x=2 is a vertical asymptote of the equation, we can only estimate the rate of change at x=2? 2a) using [f(x2)-f(x1) ] / [ (x2)-(x1)] I got -1.9998 as my final answers meaning that the instantaneous rate of change is approx -2 b) It will not because the function has a odd degree meaning that it's no symmetrical so you therefore will not have points that are equivalent? 3a) instantaneous rate of change is 0 b) using [y2-y1 / x2-x1] we get -4/3 as our average rate of change over the interval -1<= x <= 2 4a) (f+g)(x) = x + 3 b) (f+g)( 6) = 9 c) fg(x) = -6x^2-x+2 d)(f-g)(x) = 5x+1 e)(f / g)(x) = [3x+2 / -2x+1 ] therefore since our VA is 1/2 domain is {xer, x=/= 1/2} 5a) (f o g)(pi) = 1 b) ?????? I took a guess and got sin^-(sinx) 6) unsure of alot of it but this is what I had.. missing A and G altogether because I had no clue odd functions: BDE Functions with Asymptote: DH Sinusoidal Functions: BC Functions with the end behaviour..... : none Functions without any zeroes: D 7a) { x e r, x > 0 } b) ??????
 July 22nd, 2012, 11:05 AM #2 Member   Joined: Jul 2012 Posts: 50 Thanks: 0 Re: Average Rate of Change & Composite Functions http://imageshack.us/photo/my-images/215/810v.jpg/ 6 http://imageshack.us/photo/my-images/6/1113e.jpg/ 7 8 ) no clue 9a) I had no idea but I took a swing and had (f o g)(0.90x) and got 0.13(0.81x) as my final answer... probably wrong b) ? 10a) using the m = [y2-y1 / x2 - x1 ] we get [ 45 - 80 / 4 - 3 ] = -35 so the average of change between 3rd and 4th second is -35m/s b) since it hits the ground at 0 we can only use the points 24.99 and 25 in relation to this question so we simply plug this into our formula f(x2-x1) / (X2-X1) and get -249.95 m/s so APPROX -250m/s when it hits the ground 11-13) working on it but I'm really confused as to how to even do these.
 July 22nd, 2012, 03:58 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average Rate of Change & Composite Functions 1.) a) Correct. b) It is not possible, since the function is undefined for x = 2. 2.) a) To estimate the instantaneous rate of change m at x = 1, we may use: $m=\frac{f(1+h)-f(1)}{h}=\frac{$$(1+h)^4-2(1+h)^3+1$$-$$1^4-2\cdot1^3+1$$}{h}=$ $\frac{1+4h+6h^2+4h^3+h^4-2-6h-6h^2-2h^3+1-1+2-1}{h}=\frac{h^4+2h^3-2h}{h}=h^3+2h^2-2$ As h approaches zero, then m approaches -2. b) Only linear functions have a constant instantaneous rate of change, i.e., they do not depend on x. 3.) Correct. a) The instantaneous rate of change at x = -1 is the slope of the given tangent line, which is zero. b) The average rate of change over the given interval is the slope of the given secant line, which appears to be $-\frac{4}{3}$. 4.) Correct. 5.) a) Correct. b) For any function, $f$$f^{\small{-1}}(x)$$=x$. 6.) For a function to be odd, we require $f(-x)=-f(x)$. This is true for B, D, E. Functions with vertical asymptotes are rational where for some value of x the denominator is zero, also logarithmic functions, and exponential functions have horizontal asymptotes. Thus, we have D, G, H. E and F are the sinusoidal functions. For the end behavior, we find B, C, G, H. The functions with no zeros are A, D, H. 7.) a) x can be any real value except zero. b) We require $x^3+1>0$ hence $x>-1$. 8.) For this we require a function with a horizontal asymptote of y = 0. Two examples are: i) An exponential function of the form $P(t)=P_0\cdot a^{-t}$ where $1 ii) A rational function whose denominator has no non-negative real zeros and where the degree of the denominator is greater than that of the numerator. Sorry, gotta run now...
 July 22nd, 2012, 05:01 PM #4 Member   Joined: Jul 2012 Posts: 50 Thanks: 0 Re: Average Rate of Change & Composite Functions Come back ! But once again, thanks so much for your help Mark, I've been working on my other questions as well! for question 7A) : I got 2^ (2/x) as my answer and therefore the domain is D: {X E R, X=/= 0} Because if you plug in 0 you get math error 7B) I got log(x^3+1) and the domain would be { X e R, x< 0} because anything negative in a log function gives you an error for 13) let f(x) = 2^x let g(x) = 6x+7 and using f(g(x)) I get h(t) for 8 ) can I just say I would use a rational and logarithmic function? I'm a bit confused by your answer. Still stuck on 9) ... and 11) for 12) used (f o g)(x) to get (3x-2)^2 (g o f)(x) to get 3x^2-2 then I graphed the two functions and got this http://imageshack.us/photo/my-images/707/graphrj.png/ ( red is (3x-2)^2 and blue is 3x^2-2) from there we can see that vertex from 3x^2-2 is -2,0 and for (3x-2)^2 our vertex is 2/3, 0 . 13) I simply let f(x) = 2^x and g(x) = 6x+7 and using h(t) = f(g(x)) ... and simply subbed in my variables and got the original function of h(t)
 July 22nd, 2012, 05:50 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average Rate of Change & Composite Functions For 7b) you want to set the argument of the log function greater than zero. To get a real result, the argument must be positive. So set $x^3+1>0$ and solve for x, as I did above. For 8, two simple examples are an exponential function of the type I gave or a rational function as I described, and also whose numerator has no non-negative zeros. For example: $P(t)=P_0\cdot2^{-t}$ $P(t)=P_0\cdot\frac{1}{x+1}$ I will return shortly...
 July 22nd, 2012, 06:21 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average Rate of Change & Composite Functions 9.) a) Here we want the composite function: $HST(x)=f$$g(x)$$=0.13$$0.9x$$=0.117x$ b) Here we want: $HST(39.99)=0.117(39.99)\approx4.68$ 10.) a) Correct. b) Incorrect. I used energy considerations to quickly check and got 50 m/s, then used differential calculus to confirm this answer. Setting $h(t)=0$ we find $t=5$. So we then can use: $v(5)\approx\frac{h(5+h)-h(5)}{h}=\frac{$$125-5(5+h)^2$$-$$125-5^2$$}{h}=\frac{5$$-25-10h-h^2+25$$}{h}=\frac{-5h$$10+h$$}{h}=-5(10+h)$ As h goes to zero, then $v(t)=-50$. We watke the absolute value of velocity to get the speed of 50 m/s. I will come back later for the remaining 3 questions.
 July 22nd, 2012, 06:43 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average Rate of Change & Composite Functions 11.) We could begin with: $f(x)=-a(x-2)^2+b$ where a is a positive real number and b is any real number. This satisfies all three conditions. 12.) Almost correct. $f$$g(x)$$=(3x-2)^2=9$$x-\frac{2}{3}$$^2$ vertex at $$$\frac{2}{3},0$$$ $g$$f(x)$$=3x^2-2$ vertex at $$$0,-2$$$ 13.) Correct.
 July 24th, 2012, 05:39 PM #8 Member   Joined: Jul 2012 Posts: 50 Thanks: 0 Re: Average Rate of Change & Composite Functions Mark, for 5b I have as my final answer: =g^-1 (g(x)) =g^-1(sinx) =sin^ (-sinx) = x is this correct?
 July 24th, 2012, 06:14 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average Rate of Change & Composite Functions Your final result is correct, but be careful with the notation. Note that: $f^{\small{-1}}(x)\ne$$f(x)$$^{\small{-1}}$ The left side represents the functional inverse while the right side represents the multiplicative inverse. It is an unfortunate and confusing notation, but one we are given to use. You should state: $g^{\small{-1}}$$g(x)$$=\sin^{\small{-1}}$$\sin(x)$$=x$ where we have restricted the sine function's domain to a one-to-one interval containing x.

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