What is the inverse of f(x)=e^x?
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
238752
ln and e^x are inverse functions. So you can use them to solve equations.\n\nA looks like it's already solved, so just use a calculator and find the ln of 0.0093, which is -4.67774\n\nFor equation B, take the ln of both sides.\n\nln(e^x) = ln(62.5)\n\nThe ln(e^x) cancels each other out.\n\nx = ln(62.5)\n\nx = 4.13517
985432
well...ln(x) does not cancel e^x....but...\n\ngiven e^x, if we take the natural log of e^x then we have ln(e^x), well this means that we have an exponent...we recall that we can pull the exponent down infront of a logrithm...so we now have xln(e)\n\nah! the ln(e) is 1, we recall that ln's are base e and if we take the logrithm of the base we get 1. so we have x*1=x\n\nso...likewise...lnx^2 is 2ln(x)...and thats all we can do\n\nhope this helps\n\nmatttlocke
760867
f(x) = (x+3)(x^2 + 16)
809518
Peter got b and c right, but a is wrong.\n\nG^-1 (x) does not mean square root - it means the inverse of the function.\n\nThe inverse of a function is such that f(f-1(x)) = f-1(f(x)) = x.\n\nTo find the inverse of a function, do the following:\nG(x) = 2x + 4\nlet y = G(x), then solve so that x = a function of y.\ny = 2x + 4\ny - 4 = 2x\n(y - 4)/2 = x = G-1(x) (solution)
1070447
It's just like the other one.\nf'(x) = (x^3)'e^(6x^2) + (x^3)(e^(6x^2))'\nf'(x) = (3x^2)e^(6x^2) + (x^3)(12xe^(6x^2))\nf'(x) = (3x^2)e^(6x^2) + (12x^4)e^(6x^2)\nf'(1) = 3e^6 + 12e^6\nf'(1) = 15e^6\nf'(1) = 6051.432
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
909269
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
999317
(1) Simple inverse is log X as long as log is defined for the specific value of X, except that it's also not defined at X=1 since log(1) = 0.\n(2) Inverse function: If Y = 1/logX, then X = exp(1/Y) assuming natural logarithms. This is defined except at Y=0, which can't hold for finite X.\n\nThe proofs of both these are trivial. They are obvious.
316594
The anti-derivative of x^(2)*e^(x)^(3) is (1/3)*e^(x)^(3)\n\nYou can always check by taking the derivative of your answer.
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
609321
One is the inverse of the other
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
278447
answer No 1 is bullshit. Answer No 2, I couldn't even understand what he/she wants to say.\nBut answer 3 is correct:\nint(e^(x-y)=int(e^-y e^x) = e^-y int(e^x) = e^-y e^x +c=\ne^(x-y) +c
752287
I'd suggest that you draw a picture. The area 'under' the inverse function can also be seen as the area to the *left* of the graph of y=f(x), i.e. x=f^(-1)(y).\n\nThe area under y=f(x) consists of a rectangle bounded by y=2, y=0, x=7 and x=11, and the area under y=f(x) and above y=2 from x=7 to x=11. By subtraction, this second area is 6.787151.\n\nNow the area to the left of y=f(x) consists of a rectangle bounded by y=2, y=5, x=0 and x=7 and the second area above. This gives a total of 21+6.787151=27.787151. This is the answer you seek.
879979
f(x)=(x^2-2x)(x^2+2.44x+1.4884)\n =(x^4+2.44x^3+1.4884x^2-2x^3-4.88x^2- 2.9768x)\n =(x^4 + 0.44x^3 - 3.3916x^2 - 2.9768x)\n\nF(x)=(x^5)/5+(0.44x^4)/4-(3.3916x^3)/3-             (2.9768x^2)/2+c\n\nam i really asked a Q?