its an mathmatic question.. If f(x)= sx when x does not equal to zero, and f(x)=1 when x=0?
926148
What is s?\n If it is just a number then it is just a straight line. And the intercepts are  (0,1) and (0,0).
197379
x is anything less than 1.\n4 x1=4 so 1 cannot be the answer (unless the < had a line underneath it)\n4x0=0 which is less than 4\n4x -2= -8
168972
Your question does not make sense. x cannot be less than or equal to 0 and less than or equal to 7 unless it is less than or equal to 0, which I do not think you mean. \n\nI think you mean 0<=x<=7\n\nThe rest of the question makes no sense. \n\nAnd it really smells like a homework problem. Try thinking for a change.
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
951210
ok...well 1/x becomes basically infinity as x gets really close to zero...so (1/x)/x woud be infinity/something really close to 0 which would make the value infinity\n\nso we have\n\ne-(1-x)^infinity\nwell...thats ok\n\ne-(1 - (basically zero))^infinity is\ne-1^infinity\nwhich is\ne-1\n\nso the limit as x approachs zero from the right is e-1 which is approximately 1.71828\n\nhope this helps\nmatttlocke
365867
2x - (7 + x) <= x\n2x - 7 - x <= x\nx - 7 <= x\n-7 <= 0\n\nSince we're left with a statement with no variables, and the statement is true, x = all real numbers.
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
912068
suppose f(x)=x has no real root.\n\nDon't you mean f(x) = 0 has no real root?\n\n\n1) If f(x) = 0 has no real root then ot means that there is no R (R real) such that f(R) = 0 ie f(R) = c where c is real for all R\n\nf(f(R)) = f(c) ≠0 as f(Any Real number) ≠0 from the above statement\n\n2) Suppose that f(f(x)) has a real solution.\nf(f(s)) = 0 where s is real\nNow f(s) = k, a real number\nSo f(k) = 0 which contradicts the original statement that f(x) has no real solutions
821547
f(x)=4x^2 + 9x - 2\n\nhere, this means the function (f) of x=4x^2 + 9x - 2\n\nso if there was f(2)=4x^2 + 9x - 2  that would mean solve the function if x=2
905755
you must include x=0.
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Counterexample: 1\n\nJay's proof is false because x(x-1) = 0 just says that x could be either 1 or 0 to satisfy that equation. x was 1 in that case. It does not mean that 1=0.\n\nJZ-LP's proof is false because you can't say let 1=0.\n\nBianca is just plain wrong.
847176
f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
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Hint: If an analytic function is 0 on a set with its accumulation point, then the function is identically 0.\n\nNow, either f or g has to have this property since at each point one or the other is 0.
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Homework?\n\nSuppose f is not identically 0. Then there is an interval where f stays away from 0. The g be the step function of this interval to get a contradiction.
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B is false: Let a be any number other than zero. If a>0, 2(2a)-a > 0, and f'(2a) = 2. Conversely, if a<0, 2(2a)-a<0, and f'(2a) = -2. Thus, any value of a other than zero serves as a counterexample.\n\nA is also false, although I'm not sure your teacher knows this. She may be thinking of something like f(x)=[a, b]∫g(t) dt, which is alawys a constant function regardless of g(t), and thus its derivative is zero. However, if the free variable appears in the limits of integration, the function may be nonconstant: for instance, let f(x) = [1, x]∫1/t dt, which is equal to ln x for positive values of x, and its derivative is certainly not zero. So the correct answer here is d: both A and B are false.
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
1032037
it is any number smaller then 4, for example 3.99999 or 2, or 0.00001, BUT NOT zero itself!!! (0=0, and not 0<0)\n\nSo x<4 is WRONG, the answer is: \nx<4 AND x is not equal to 0\nBUT x CANNOT be a negative number either, as -1(-1-4)=5!!\n\nSo the final answer is: \nx<4 and x>0 --- any figure will match your condition\n\nThanks for the math!