The standard equation of an exponential function is y=ab^x...Why can't "b" ever be negative in this equation?
108073
For the standard equation, the value of y has to be defined for all values of x, AND the derivative dy/dx has to be defined for all x.\n\ny = a*b^x = a*e^(ln(b^x)) = a*e^(x*ln(b))\n\nif b < 0, then ln(b) is undefined.  So b cannot be negative.\n\ndy/dx = ln(b)*a*e^(x*ln(b)) = ln(b)*a*b^x\ndy/dx = ln(b) * y, again, requiring that b cannot be negative.
811215
If the value in an absolute value is positive, it remains the same. If it is negative, it is multiplied by -1.\n\nEither a<x, a = x, or a > x. (I think this is called trichotomy?)\n\nIf a = x, then both sides equal 0.\n\nIf a > x, then a-x is positive, and x - a is negative, so...\na - x =  -1(x - a)\n\na - x If a > x, then x-a is positive, and a - x is negative, so...\n-1(a - x) =  x - a\nx - a = x - a\n\na - x = a - x\n\nQED
448269
The domain is all real numbers x, except for x=0.\n\nYou just have to remember that the exponential function and all polynomial functions are defined on the entire real line, that is there domain is all real numbers. Therefore, the only problem that occurs is with the operation of division by x. Since you cannot divide by 0, x can never be 0. All other values are fine though.\n\nIN RESPONSE to what urtimud said below: sure you can take e^0, but if you let x=0 you are taking e^(-1/0) which is undefined. Therefore, x=0 is not in the domain of this real valued function. Note that e is being raised to the ((x^2-1)/x) power, because your screen may have chopped off that last part.
256543
I'll assume you mean y=a e^(kx) +b e^(-kx). Then dy/dx=ak e^(kx)-bk e^(-kx). Set this to 0 and do some work to see that\ne^(2kx)=b/a, so x=[1/(2k)]ln(b/a). If b/a <0, there is no critical point. If a<0 and b<0, the critical point is a maximum, not a minumum. So, for a>0, b>0, we get the minimum occurs with the x value above. Plug in and use properties of exponents and logs to find that the minimum value is f(x)=2sqrt(ab).
83855
y = a*x^2 + b*x + c\n\nThat is, y = constant times x squared plus constant times x plus a third constant. The solution is Eqn 5 on the link below.\n\nIf the stuff under the radical in the solution is negative then you will have an imaginary number on your hands.
721733
if y is always increasing then y' is always >=0 (1) \ny'=3ax^2+2bx+c \ndelta': D=b^2-3ac (2)\nwe also have a>0 (3)\n(1),(2)&(3)--> D <=0-->b^2-3ac <=0-->b^2<=3ac (qed)\nNote: we have b^2 <=3ac, not b^2 < 3ac,for example : function y=(x-1)^3
744013
This is not common.\n\nIn general, you're looking for x^y = y^x.\n\nThis is equivalent to x^(y/x) = y, which in turn is equivalent to x^(1/x)=y^(1/y).\n\nx^(1/x) has a maximum at e^(1/e) = 1.44466....  The function is increasing before that point and decreasing after.  As a result, for each number there is another on the opposite side with the same value.  2^(1/2) = 4^(1/4) <--> 2^4 = 4^2 is the only example of this in the integers.
1125670
Let x = (-2)^(-2)^(-2)^(-2)^(-2)^(-2)^...\n\n(-2)^(-2)^(-2)^(-2)^(-2)^(-2)^... = (-2)^(-2)^(-2)^(-2)^(-2)^(-2)^...\nx = (-2)^x\nln(x) = x * ln(-2)\n\nNot a real number, that's for certain.  I have to brush up on my complex logarithmic functions.... :-(
498337
from the first equation, x=(-a-b)/2\n\nsubstituting x in the 2nd:\n((-a-b)/2+a)^2=((-a-b)/2+b)^2\n((a-b)/2)^2=((b-a)/2)^2\n((a-b)^2)/4=((b-a)^2)/4\n(a-b)^2=(b-a)^2\na^2-2ab+b^2=a^2-2ab+b^2\n\nI disagree that this is not possible if a=b:\n(a+b)(a-b)+2x(a-b)=0\n\nif a=b, then:\n2a(0)+2x(0)=0\n2b(0)+2x(0)=0\n0+0=0\n\nHow does that not work? I think you are trying to say you can't divide by zero, but it is possible under both cases.
797160
Your function y=4*x^2*e^(5x) is defined for any x. Meanwhile x^2>=0 & e^(5x)>0, thus y>=0 for any x. Now y’=4*(2x+5x^2)*e^(5x), local extrema being at y’=0. Or x1=0, x2=-0.4\nHence and “above” x1=0 means local min & y(0)=0, so x2=-0.4 means local max, \nLim[for x->-infinity]=0 & Lim[for x->+infinity]=+infinity. Does it help you Xu?
492321
The inverse function of f(x) = e^x is ln(x) because \nln(e^x) = x.  \nWhen x = 1, ln(e^1)=1 ; When x = e, ln(e^e) = e\nThe equation in exponential form is:\ne^ln(e^x) = e^x\n\nIf you have a function g(x) which is the inverse function of f(x), then g(f(x)) = x. For example, arcsine(x) is the inverse function sine(x) and arcsine(sine(x))= x
802801
i think -1; y=mx+b, "b" is the 3 and "m"(slope) is negative one.
150881
There are several reasons. One of the most basic ones is that this representation provides a "polar" representation of complex numbers. Any complex number z=a+ib can be written as\n\nz= r exp(i theta)\nr= sqrt(a^2+b^2)\ntheta = atan(b/a)\n\nThis helps because, for example, a product of z1 and z2 would give\n\nz1 z2 = R exp(i phi)= r1*r2 exp (theta1 + theta2)\nR=r1*r2\nPhi = theta1 + theta2\n\nSo multiplying by z2 "stretches" z1 by amount r2 and "rotates" z1 by theta2. Such interpretations are convenient in many instances. Also, exponential functions are easier to deal with than regular a+ib type expressions in other cases.
1019179
Yes, negative x positive always equals negative.  \n        negative x negative always equals negative\n        positive x positive always equals positive
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!
863581
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
362293
The logarithm (see attached article 1) of a number x in base b is the number n such that b^n = x. It is usually written as log_b(x) = n (b in a subscript).\n\nFor example, log (100) = log_10 (100) = 2, because 10^2 = 100.\n\nThere is one special case - natural logarithm (in base e =~ 2.7182818). It has some properties, which are very useful in calculus. They are written as ln (x).\n\n\nAntilogarithm is a synonym of exponentiation or power.\nThis is an inverse function to logarithm.\nIf log_b(x) = y, then b^y=x.\nFor integer x, b^x mean multiplication b*b*b*...*b (x times).\nFor all real x, b^x are defined in the second attached article, part "Exponentiation with real exponents".\n\nHope it helps.