Find the value if (f o g)' at x=1.?
f(u)=cot pi*x/10 Using the chain rule f'(u) = -(pi/10)csc^2(pi*x/10)
How to find the value of (f o g)' at the given value of x?
Use the chain rule. d/dx [f(u)] = f'(u) * du/dx f'(u) = d/dx [tan(πu/2)] = sec²(πu/2) * d/du [πu/2] = (π/2) sec²(πu/2) du/dx = d/dx [x²] = 2x (f o g)' = (π/2) sec²(πx² / 2) * (2x) = πx sec²(πx²/2) Then just plug in x = 2. http://www.wolframalpha.com/input/?i=d%2...
Find the value of x for which f(x) = g(x) ?
set f(x) equal to g(x) x^2 = x + 2 to get an eqation that you can plug into the quadratic formula, you need to get everything on one side so that you eqation is in this form... ax^2 + bx + c = 0 so in your case, add -x - 2 to each side, and you get x^2 - x - 2 = 0 if you don't know the quadratic formula, it is {-b +- sqrt[b^2 - 4(a)(c)]} / [2(a)] in this case.... a=1, b= -1, c= -2. plug in the numbers and you get {1 +- sqrt[-1^2 - 4(1)(-2)]} / [2(1)] [1 +- sqrt(1 + 8)] / 2 (1 +- sqrt9) / 2 (1 +- 3) / 2 1 + 3 / 2 = 4/2 = 2 1 - 3 / 2 = -2/2 = -1 so your answer is x = {2, -1}
Find the value of (f o g)' at the given value of x?
f(g(x)) = [g(x)]³ f'(g(x)) = 3[g(x)]²g'(x) g'(x) = [(x-2) - (x+4)]/(x-2)² = -6/(x-2)² g(x) = 9/3 = 3 g'(x) = -2/3 f'(g(x)) = 3[3]²(-2/3) = -2*9 = - 18 ...
What is the next value of 4 D 7 G 10 J 13?
The letter value is equivalent to the numerical value. The numerical value comes before the alphabetical value, so the letter comes after the number. The numerical and alphabetical values are increasing by a steady rate of 3 or C in the sequence.4=D7=G10=J13=?D+C=G or 7G+C= J or 10J+C=M or 13Let's continue the sequence, shall we?M+C=P or 16P+C=S or 19S+C= V or 22V+C=Y or 25BonusY+C= Z+B or 28Or you can use your imagination.
If f(x) =3x^2 and g(x) =x-1, how do you find f(g(x)) and g(f(x))?
JaX a boy fooling around people he stumble with. He would ask for your birthdate and would put on your back a sticker with your birthdate.F is an unrully boy who would get your sticker, put it to the power of 3 and multiply by 3. F = 3*(x*x*x)So as I was born on October the third. JaX had stuck a 3 on me and F replaced it by a 81 G is more assertive. He just add one to your sticker number.When G saw me, he replaced it by an 82 .So there is a jaX who assigns you a number XIf you see G first he will stick a x+1 and if you see F after G he would stick a 3(x+1)^3So the sequence is jaX G and FAnd the stickers x x+1 3(x+1)^3. But if you see JaX and F and then GThe sequence is x 3x^3 3x^3 +1
How can I find the minimum value of the function [math]f(x)=x^x-10^x[/math]?
If you just want a quick and dirty answer without showing any work, type the equation into a graphing calculator, experiment to find window settings that show the equation, then use the Calc-Tools to find the minimum.I did that here:If I want to actually do the work, I’d refer to my reference sheets, find the common derivatives, and write down the derivative of your function:Here’s the reference sheet that i use, but if you want to print it, go to their website, download the PDF files (there are six in total), and print them.Once you do all the work, you will have:f(x) = [math]x^x + 10^x[/math]f’(x) = [math]x^x (ln(x)+1) + 10^x ln(10)[/math]Set that equation to zero, solve for x, and you will find your X value, then you can plug that value into your original equation to find the minimum.I’ll let you do the rest.EDIT:Information about Newton-Raphson method supplied by ZilleplusOne way would be to use Newton Raphson:The equation f’(x)=0 is an non-linear equation, an easy way to solve this is with the Newton-Raphson method. This is an iterative method, each time you execute the iteration you get a better solution.The general formula of one iteration:[math]x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}[/math]g(x) is the function of which we want the roots. (as in find x when g(x)=0).It easy to derive it from the Taylor expansion, google this is you want to know more.If we use the rules of calculus again.[math]f"(x)=x^{x-1}+x^x(ln(x)+1)^2+10^x+log^2(10)[/math]now set g(x)=f’(x) and g’(x)=f”(x) and we get:[math]x_{n+1} = x_n - \frac{f'(x_n)}{f"(x_n)}[/math]Do this in a loop till you have the amount of digits that you want. (use Python/Matlab or [enter your favorite programming language])One important thing to remember is that this converges very fast only when its near the solution. You have to chose your start value yourself, try it out with your favorite number.The negative thing about the Newton-Raphson method is that you need the derivative which is not always possible, but this is not an issue here.***** Thanks to Zilleplus for the instructions on how to use the Newton-Raphson method. I’ll have to study this some more.
How to find fog from this table?
You shouldn't be looking at a table, you're more likely to find fog near a swamp or early in the morning... OK, well the first is easy, I HATE this notation...but then I know people that don't like the notation I use (it's not mine, there are basically just two ways to write these): (3f + 2g)(1) = 2 * f(1) + 2 * g(1) and (f o g)(-1) = f(g(-1)) For the first, we will need to find f(1) and g(1) So looks like we have (from the table): f(-3) = 10 g(-3) = -1 f(-2) = 20 g(-2) = -2 f(-1) = 30 g(-1) = -3 f(0) = 40 g(0) = -4 f(1) = 50 g(1) = -5 f(2) = 60 g(2) = -6 f(3) = 70 g(3) = -7 f(4) = 80 g(4) = -8 f(5) = 90 g(5) = -9 So the only one we need for the first problem is: f(1) = 50 and g(1) = -5 So... 3 * f(1) + 2 * g(1) = 3 * 50 + 2 * -5 = 150 - 10 = 140 (3f + 2g)(1) = 140 For the second, first we need g(-1): g(-1) = -3 --> we want f(g(-1)) --> plug in g(-1) f(-3) --> look up in your table f(-3) = 10 so... f(g(-1)) = 10 or (f o g)(-1) = 10
Let [math]f(x)=3x^2 -1, \; g(x)=x+2.[/math] How do I find [math]f(g(x))[/math] and [math]g(f(x))[/math]?
The notation f(g(x)) is read as f of g of x. It means wherever there is x in the function of f, it is replaced with the function g(x).To find f(g(x));f(x) = 3(x^2) - 1g(x) = x + 2f(g(x)) = 3(x+2)^2 - 1 ----------------------replacing x with g(x) = 3(x^2) + 12x + 11g(f(x)) = [3(x^2) - 1] + 2 -------------------replacing x with f(x) = 3(x^2) + 2