Why is the following false?  If f(x) = x to 2nd power - 1, then f(x+h) = x to 2nd power + h -1 Correct it?
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
766600
You have to use the quotient rule for this problem:\n\nFor any function f(x) defined by, f(x) = g(x)/h(x),\n\nf'(x) = [g'(x)h(x) - h'(x)g(x)]/[h^2(x)]\n\nIn this problem, we have:\n\ng(x) = xe^-x\nh(x) = x^3+x\n\nTherefore, \n\ng'(x) = xe^-x(-1) + e^-x(1) [you have to use product and chain rule for thie derivative]\n\nSimplfying, you get e^-x(1-x)\n\nh'(x) = 3x^2 + 1 (power rule)\n\nNow that we have all parts of the derivative, just plug in:\n\nf'(x) = [e^-x(1-x)(x^3+x) - (3x^2+1)(xe^-x)]/(x^3+x)^2\n\nNow just simplify this ugly mess =P\n\n---------\n\nHope this helps
918705
Don't you see the pattern?\nIt's right in front of your eyes.  Let g(x) be what is in the parentheses.  So g(x) = x+2\nNow replace x+2 with just x, and that will be f(x).  \nSo f(x) = x^3 - 5x^2 + 3x - 1\nAgain, this is assuming that h(x) = f(g(x)) (based on your other problems)
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
948694
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x). \n\nIf f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1)  + C ... we need to add an arbitrary constant.\n\nIn your case we have:  f(x) = x^1  and  F(x) = (x^2)/2 + C
534496
I think this says\n\n(x^3 - x)/(x^2 - 1)\n\nIf that's the case, then\n\n(x^3 - x)/(x^2 - 1) = x(x^2 - 1)/(x^2 - 1)\n= x
1071898
it's x, assuming that x is zero or postive. \n\nif x is negative, it is either undefined, \nor if you use complex numbers, it is |x|=-x.
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
783756
in standard form yes, but thats not the whole answer, the answer would be x to the 2 power and -1
764395
Hey, firstly I need this corrected: f(x)= Ix^2=5x+6I, because there should not be an equal sign after the power of 2.... I can't do this w/o knowing... if it is subtraction than I will tell you that from what I rembember about this stuff is that absolute value functions are continuous but not differentiable I think.... Because at the point, the f(c) equals the f(x) which means it is cont. but the derivative of f(x) approaching the vertex from the left side does not equal the derivative of the f(x) from the right side...whcih means it is NOT differentiable.... sorry I can't remember more buddy
574042
we need to multiply something by (x+1) so that the result would be x^2-1. but we know x^2-1 = (x-1)(x+1).\nso we have to multiply (x-1) with (x+1) to get x^2-1.
153289
The answer is: f(-x)=1-(-x-1)^1/2.\n\nIf f(x)=f(x)=1-(x-1)^1/2, then it means that f(y)=1-(y-1)^1/2. Let y=(-x), then we get f(-x)=1-(-x-1)^1/2. Note that -x must be less than -1 in order to get the answer in real number. Otherwise, it will be a complex number.
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
162293
On both equations, you have x^ but don't say what x was raised to. I'm assuming that it is x^2. Is this correct?
485166
The correct answer is d.\n\nf(1-x) will give a negative number for all positive 'x' values.\n\nbut squaring these negative number will always give a positive number i.e (-4)^2 = +16.\n\nTherefore in the equation f(x) = x^2(1-x)^2;\n\nYou can see that for all positive or negative numbers you will have a positive value. Again,\n\nLet x=5; Then 1-x = -4;\n\nFor the equation d.. 1-(-4) = 5. i.e the original number.\n\nHence, f(5) = 5^2 (-4)^2 = 25*16;\nAlso, f(-4) = (-4)^2 (5)^2 = 25*16;\n\nThus, for all values of x... Equation d is TRUE.