If f(x) is a linear function, where f(3) =2, and f(-3) = -10.?
610276
points (3,2) and (-3,-10) belong to f and the equation for f is\ny-y1=[(y2-y1)/(x2-x1)]*[x-x1]\nwhere y1=2,y2=-10,x1=3,x2=-3\n\ny-2=[(-10-2)/(-3-3)]*[x-3]\n\ny-2=[-12/-6]{x-3]\n\ny-2=2[x-3]\n\ny=2+2x-6\n\ny=2x-4\n\nso the slope of f is 2\n\nsince g is perpendicular to f the slope is -1/2\n\nand equation would be\n\ny-9=-1/2(x+4)\n\ny=9-[1/2]x-2\n\ny=-(1/2)x+7\n\nso g(x)=-(1/2)+7
791940
In order to a od function the function have to obay:\nf(x) = - f (-x) <=>\n-3x^9 + 4x^5 = -(-3(-x)^9 + 4(-x)^5) <=>\n-3x^9 + 4x^5 = 3(-x)^9 - 4(-x)^5 <=>\n-3x^9 + 4x^5 = -3x^9 + 4x^5 ( True)\nthe function is od.
622612
Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
650997
f(3(x - 4)) \n= f(3x - 12) \n= (3x - 12 + 1)^2 - 4 (3x  - 12)\n= (3x - 11)^2 - 12x - 48
578027
f(x)= 7x-6\nf(0)= 7(0)-6= 0-6= -6\nf(1)= 7(1)-6= 7-6= 1\nf(-2)= 7(-2)-6= -14-6= -14+(-6)= -20\n\nJust pretty much substitute for x with each value.  Hope this was of some help to you.
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
1086560
f(-2) = 5. What's so difficult about this question?
572493
=x(x^3-2)\n--> x=0 or x=2^(1/3) or -2^(1/3)
150385
f(x)=Int(f'(y), 0, x)-f(0) => yes, they're opposite operation, but if x has a given value, e.g. x=2, 3...no, they're not, b/c in that case, the RHS of the previous egality is a real number, not a function
958467
Replace x with 2:\nf(2)=2^3-4(2)^2+14(2)-20\nf(2)=8-16+28-20\nf(2)=-8+8\nf(2)=0
655260
2=-1=x\n2+1=x\nthere fore x =3
352533
Generally speaking, you call "linear" a function F (or an operator whatsoever) which satisfies the following conditions:\n\n1) F(a*X)=a*F(X) for every real number a (X is the argument of the function, whatever it be: it can be a number, a vector, a complex, a function... it doesn't matter)\n\n2) F(X+Y) = F(X)+F(Y) for every X and Y which can be taken as an argument of the function or the operator.\n\nThese two conditions can be sinthesized in the following single one:\n\nCONDITION FOR LINEARITY:\nF(a*X+b*Y) = a*F(X) + b*F(Y) for every a and b reals and for every X and Y in the domain of the function.\n\n----\n\nIn the ring of the polynomials of n variables x1, x2 ... xn, it is easily verified that the only polynomials which can be defined "linear" according to the above conditions are those in which the variables appear at the first degree only, so they are in the form:\n\nP(n) = a1*x1 + a2*x2 + ... + an*xn + a(n+1)\n\nLinear equations in n variables are obtained by setting a linear polynomial of the above type equal to 0.\n\nNon-linear equations are all other possible kinds of equations: the ones obteined by setting to 0 non-linear polynomials (you call them algebraic non-linear equations), or the ones obtained by setting ot 0 expressions which involve logarithms, exponentials etc (in that case you are dealing with transcendant equations)
914146
For x < 1 \nf(x) = |x-1| + 2\nWhen x → 1-, f(x) →2\nWhen x < 1, |x-1| + 2 = 1 - x + 2 = 3 - x. So slope is -1.\n\nFor x ≥ 1\nFor continuity f(1) = 2.\nFor differentiability f'(1) = -1\n\nSo a + b = 2\n2a + b = -1\n\nThus a = -3, b = 5\n\nf(x) = |x-1| + 2, for x < 1 \n..... = 5x - 3x², for x ≥ 1\nis continuous and differentiable for all x
3588
You've answered your own question.\nNo, f(x) is not defined at x=3 because the denominator would equal 0.  It doesn't matter what the numerator is.\n\nAnd btw, the expression simplifies to f(x)=x+3, where x<>3, \nbecause (x^2)-9=(x+3)*(x-3).
781968
I'm not sure if your question meant f(x)=2x(x-1)(x-4) or f(x)=x(x-1)(x-4)^2, so I will answer both.\n\n1) f(x)=2x(x-1)(x-4)\n(0, 1) U (4, ∞)\n\n2) f(x)=x(x-1)(x-4)^2\n(-∞, 0) U (1, 4) U (4, ∞)
1055950
Yes, linear functions have limits.\n\nFor example, the limit of 3x+2 as x approaches 4 is 14. Here's how to prove it:\n\nFor every e>0, choose d=e/3. Then, if 0<|x-4|<d, we have\n|(3x+2)-14|=|3x-12|=3|x-4|\n<3d=e.\n\nFor this situation, we say that the limit of a function f(x) as x p&#92;approaches a is L is for every tolerance e>0, there is a tolerance d>0 so that whenever 0<|x-a|<d, we have\n|f(x)-L|<e.