I have a question about mathamatics. How do I solve questions like f(x). Exp. f(x)=9     f(x)=x=5?
109954
get "x" by it's selef
909799
gopal's solution:\n\nf(x)=x^3-6x^2+9x+2\nf'(x)=3x^2-12x+9=0 for extremas\n3(x^2-4x+3)=0\n3(x-3)(x-1)=0\nx=3 or 1\nf''(x)=6x-12\nat x=1 6(1)-12=-ve so max\nat x=3 6(3)-12=+ve so min\nso the absolute maximum is at x=1 and\n=6\nabsolute min is at x=3 and is=2  <------ However still need to check endpoints of the interval [-1,4] as these turnng points are actually LOCAL maxima and minima!\n\nSo f(-1) = -1 - 6 - 9 + 2 = -14 < 2\nSo ABSOLUTE minimum = -14 at x = -1\n\nAlso f(4) = 64 - 96 + 36 + 2\n= 6\nSo ABSOLUTE maximum = 6 and occurs at both x = 1 AND at x = 4
937835
f(x) = ax^3 + bx^2 + cx + d\nFrom the given points, you know that\n(3,3) implies f(3) = 3\n(5,1) implies f(5) = 1 and\n(4,2) implies f(4) = 2\nUsing those three pieces of information, you have\nf(3) = a(3^3) + b(3^2) + c(3) + d = 3\nf(5) = a(5^3) + b(5^2) + c(5) + d = 1\nf(4) = a(4^3) + b(4^2) + c(4) + d = 2\nSimplify each one\n27a + 9b + 3c + d = 3\n125a + 25b + 5c + d = 1\n64a + 16b + 4c + d = 2\nNow you have four unknowns but only three equations.  \nWe can find a fourth equation from the fact that (3,3) is a relative extrema.  Since (3,3) is a relative extrema, that means f'(3) = 0\nFind f'(x)\nf'(x) = 3ax^2 + 2bx + c\nReplace x with 3 and set it equal to zero.\nf'(3) = 3a(3^2) + 2b(3) + c = 0\nor 27a + 6b + c = 0\nWe now have four equations and four unknowns.\nSolving this system of equations you get that \na = 1/2\nb = -6\nc = 45/2 and\nd = -24\nThat means\nf(x) = (1/2)x^3 - 6x^2 + (45/2)x - 24\nYou can verify my answer by checking that\nf(3) = 3, f(5) = 1, f(4) = 2, f'(3) = 0, f'(5) = 0 and f''(4) = 0.\nThis will guarantee that my answer is correct.
1102322
If you're in calc...\n\nf(x) = -2x^2 + 12x\n\nTake the derivative:\n\nf'(x) = -4x + 12 = -4(x - 3)\n\nWhere f'(x) = 0, the slope of f(x) is 0.\n\n0 = -4(x - 3)\nx = 3\n\nSo, at x = 3, f(x) is a maximum or minimum. Put 3 in f(x).\n\nf(3) = -2(3^2) + 12(3) = -18 + 36 = 18\nf(3) = 18\n\nGiven, that the answer is multiple choice, there's no need to prove that (3, 18) is a maximum. The answer is\n\n"maximum; 18"\n\nHope that helps!
1116623
y=x^2 - 4x - 5\n\nFunction:\n\ny= 1(x-2)^2 - 9\n\nNow it's up to you to find out the rest. I not going to tell you everything.
756998
f(g(x))= 4g(x) -5\n= 4(x/3) -5\n= 4x/3 -5\ns
759378
More concisely, f(-5)=-8, f(-3)=-2, f(-2)=1, f(1)=10.\n\nf(x) can equal anything as long as it fits these points. It seems to form a linear line as shown above...
35579
5(x^4 -9) = 5(x^2 + 3) (x^2 - 3)
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
1019687
F(n)=F(n-1)+F(n-2), where F(0) = 0 and F(1) = 1\nn= 2 \nF(2) = F(2-1) + F(2-2)\nF(2) = F(1) + F(0) \nF(2) = 1 + 0 = 1\nn = 3\nF(3) = F(3-1) + F(3-2)\nF(3) = F(2) + F(1)\nF(3) = 1 + 1 \nF(3) = 2\nn = 4\nF(4) = F(4-1) + F(4-2)\nF(4) = F(3) + F(2)\nF(4) = 2 + 1\nF(4) = 3\nn = 5\nF(5) = F(4) + F(3)\nF(5) = 3 + 2\nF(5) = 5
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution
535346
I'm not sure if this is the whole question, but I will interpret it as if it ended with a ) in place of the [exp...\n\n9(x^3)N - 12(x^4)(N^2) - 9(x^2)N
809444
ohh ya,,i know this,,\n\n1)f(x)=((x)^2)+2\n\n2)f(x)=( ( (x-3)^(-2) ) +2)\n\nbring on more of domain and range questions,,i love those problems,,
840626
f(x)= x^5 + 5x^4 - 9x^3 - 45x^2\n\nFirst, notice you can factor out an x^2 to get:\n\nf(x) = x^2(x^3 + 5x^2 - 9x - 45)\n\nNow let's look at x^3 + 5x^2 - 9x - 45.  Try dividing this by (x+5) and see what you get.  By long division, I get x^2-9.\n\nNow notice that x^2-9 is the difference of two squares, so it can be factored into (x+3)(x-3)\n\nTherefore, f(x) factors into x^2(x+5)(x+3)(x-3)\n\nb)  The y-intercept is simply the point on the graph where x=0.  So solve f(0) to find the y-intercept.\n\nc)  Notice that x^2 > 0 for all values of x except x=0 in which case x^2 = 0.  Therefore, to examine where f(x)>0 or f(x)<0, we need only look at where (x+5)(x+3)(x-3) is >0 or <0.  {{ I think it's best to let you see if you can take it from here, but if you get stuck please feel free to email me and I'll give you further help if I can }}\n\n~ ♥ ~
717628
f(x) = -x^3 - x^2 + x - 9\nf(-1) means replace x with -1\nf(-1) = -(-1)^3 - (-1)^2 + (-1) - 9\nf(-1) = -(-1) - (1) - 1 - 9\nf(-1) = 1 - 1 - 1 - 9\nf(-1) = -10\n\nf(0) means replace x with 0\nf(0) = -(0)^3 - (0)^2 - 0 - 9\nf(0) = 0 - 0 - 0 - 9\nf(0) = -9\n\nf(2) means replace x with 2.\nTry this one on your own.
964521
1. I think there's more information needed here.  There are several functions of x such that an x value of 4 will produce a function value of -5 (f(x) = x -9, f(x) = 2x -13, f(x) = x^2 - 5x -1, etc.).\n\n2. I assume that your original function is f(x) = 1/(x^2 - 2x - 8), not f(x) = (1/x^2) - 2x - 8.  If that assumption is incorrect, my answer will be wrong.\n\nThe only way 1/(x^2 - 2x - 8) can be negative is if (x^2 - 2x - 8) is negative.  Thus, you should set up a new inequality like this:\nx^2 - 2x - 8 > 0\n\nFactor:\n(x - 4)(x + 2) < 0\n\nNow, we know that when two numbers are multiplied together, the only way you end up with a negative is if one of the numbers is negative and the other is positive.  Thus, you need to figure out where that will be true in this case\n(x - 4) is positive for all numbers > 4 and negative for all numbers < 4\n(x + 2) is positive for all numbers > -2 and negative for all numbers < -2\n\nThe easiest way to figure out your interval would be to draw this out on a number line, but the answer is -2 < x < 4.\n\n3. 2^2 = 4.  Therefore, 4^(2x + 1) = 2^2^(2x + 1) = 2^(4x +2)\n2^(4x + 2) = 2^(3x + 6)\n4x + 2 = 3x + 6\nx = 4
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],