Write the taylor polynomial with center zero and of dgree 4 for the function f(x)=e^-x?
404885
1 - x + x^2/2 - x^3/6 + x^4/24
875938
5\n\ntake the highest power in the polynomial, that exponent is your leading exponent, and is used as your "n"\n\nso 5 zeros.
852324
We have f(x) = (x)^(7x) = e^(7x*ln(x)). So, f'(x) = e^(7x*ln(x)) * d/dx(7x*ln(x) = . (x)^(7x) *7 *(x *1/x + ln(x) = 7* (1+lnx) * (x)^(7x)
755844
1 correct \n\n2 ∫x/e^x dx \n= ∫x*e(-x) dx\n= -xe^(-x) - ∫-e^(-x). 1.dx\n= -xe^(-x) - e^(-x) + c\n\nSo∫[x = 1 to ∞] x/e^x dx = lim t →∞ (-te^(-t) - e^(-t)) - (-1/e - 1/e)\n= 2/e (NB lim t →∞ (-te^(-t)) = o by d'Hopitals rule)\n\n3 correct\n\n4. Use (n² + 1)/[e^n * (n + 1)²] \n=  [(n² + 1)² - 2n]/[e^n * (n + 1)²]\n= 1/e^n - 2n/e^n(n + 1)² \nNow 1/e^n < 1/3^n and  2n/e^n(n + 1)² < 2n/3^n*n² \n= 2/n*3^n \nSo both parts converge so the difference converges\n\nNOTE: It is many, many years since I have done these BUT I am confident that I am correct ... Maybe you can tell me if that is so!
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
951729
See this site for details of lagrange polynomials, then the program shouldn't be hard to write:\n\nhttp://en.wikipedia.org/wiki/Lagrange_interpolation
736107
If u = e^x, then 1/u = 1/e^x = e^(-x)\n\n(u - 1/u) / (u + 1/u) = -t\n\nMultiply both sides by (u + 1/u)\n\nu - 1/u = -t (u + 1/u) = -ut - t/u\n\nMultiply both sides by u\n\nu^2 - 1 = -u^2t - t\nu^2 + u^2 t = u^2(1 + t) = 1- t\nu^2 = (1 - t)/(1 + t)\nu = +/-sqrt[(1-t)/(1+t)]\n\nRe-substitute u=e^x:\n\ne^x = +/-sqrt[(1-t)/(1+t)]\n\nTake ln of both sides:\n\nx = ln[sqrt[(1-t)/(1+t)]\n\nSorry I screwed that up the first time.  I need sleep.  Thanks for getting it right, Sidarth.
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
311134
f(x) = 1/(x+5)(x+2)\n\nso -5 and -2 are restricted from the domain since then we would have the denominator = 0 and the function undefined.
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
839617
i think you meant F(0)=0\nand f(t)=6sec^2(t)-9t^3\nso\nF(t)=6tan(t)-(9/4) t^4+c\nF(0)=c=0\nF(t)=6tan(t)-(9/4)t^4
669001
e^f(x) = x is the best answer i can give you.
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
156051
first let T =X^2\nthan we get a second degree equation\n2T^2-7T-4=0\nso you can solv it easyer
275631
One way to see this is to write x^(1-x) as exp((1-x)ln x) and use the chain rule. Only one person has the correct answer, art_colls. The power rule *does not apply* since the exponent is a function also!
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2