what is true about this function?
816980
1 is false. The function is continuous.\n2 is false; see above.\n3 is true.  Hence, 4 is false, and so is 5.\nYou are right about the non-differentiability at x=3.  A function cannot be called differentiable if its slope at x- does not approach its slope at x+ as the interval approaches zero.
779526
How are you even sure that such a function exists? the numbers look very random.
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
952602
This is what always helps me.\n\nA function is a machine.\n\nThe domain is everything you can put into the machine so that it doesn't break.\n\nThe range is everything you can get out of the machine.\n\nIn the case of your function, there's only one possible value of x that will "break" the function, in the sense that you'd get an infinite answer.  That value is x=0, because then you'd have 9/0 + 7, and 9/0 is infinity.\n\nSo the range is {x | x is not equal to 0}.\n\nAssuming that you meant 9/x + 7, rather than 9/(x+7).  If you meant the latter, x cannot be equal to -7.
733792
Yes it is a function because by the verticle line test it only intersects that line at 1 point.  \n\n\n It is a pretty simple test.  if a verticle line hits the graph at more than 1 point it is not a function.
497029
No, because the limit as x->2- is undefined.  The limit exists for x->2+.
863364
Not sure if I understood your question. maybe you meant a series of functions. Suppose that, for each n you have a function f_n and, for every x in the domain of the functions f_n, you define S(x) = f_1(x) + ....f_n(x) +...provided this series converges. Then, you have a series of functions, since for every x S(x) is given by the limi of a series.\n\nFor example, consider s(x) = 1 + x  x^2/(2!) ....+ x^n(n!)....|This is a power series, in this case it converges for every real x. This is the definitioj of the exponencial function e^x
888635
Let f(x) = 3x^4 - 25 + 5/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nBecause f(x) is a polynomial function (of degree 4) on x, it follows that the function f is continuous on R (the set of real numbers).  So, if we take the limit on the left (and right) of x, we will get the same value for both sides. \n\nThat is, the behavior on the left (Left-hand Limit) will be: as x approaches  c from the left, we get f(x) approaches f(c) = 3c^4 - 25 + 5/4. Similarly, the behavior on the right (Right-hand Limit) will be: as x approaches c from the right, we also get f(x) approaches f(c) = 3c^4 - 25 + 5/4. \n\nTherefore, both the left and right-hand limit yields to f(c), i.e. as x approaches c, we always obtain f(c).\n\nHence, both left and right-hand behavior of f are the same for all real valued x.
695915
Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
71723
You just plugin one function into another function where there's an X.\n\ng(f) means use function G and plugin function F to replace the X\n\n1.  g(f)=3-2x f(g)=2(3-x)=6-2x\nI just replaced 2x with x for g(f) and 3-x for x for f(g)\n\nUse this method to figure out the rest.\n\nNotice that it says "x is not equal to whatever".\nThis just states that it will be undefined if it is that number and for now, should be ignored as all problems will have real answers.
872378
It is definitely convergent. the function n^ is of exponential order, defined as any function f(x) for which\nlimit x--->0 f(x)e^-x=0
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.
437573
Functions that maps x to x\nWhere x is an element of the function
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
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"
378109
Think about what the inverse does. i.e. f(3) * f^-1(3) = 3 better be true.  Further, f(f^-1(3)) = 3. From what you are given, does f(1/5) = 3?    I can't tell. Did you type the question completely?\n\nSecond question: g(3) = g(5).  This implies it g is not a injective (1-1) function.  Can a non-injective function have an inverse?  Say g(3) = g(5) = 100 then what does g^-1(100) equal?  3 or 5?
900593
y=f(x)= a*x*e^(b*(x^2)),\n y’=a*[e^(b*(x^2))+x*b*2*x*e^(b*(x^2))],\n y’=a*[1+2*b*x^2]*e^(b*(x^2))=0,\nThe value of e^(b*(x^2)) is always >0, also ‘a’ is not =0. \nthen 1+2*b*x^2=0, while x=3, b=-1/(2*3*3)=-1/18\n& 4=a*3*e^(-3*3/18), hence a=(4/3)*sqrt(e)\nis it really MAX?\ny’(0)=a>0 means increasing function, thus YES!