The average value of f(x) = 2x^3 - 5x on the interval [-4, 4] is?
957499
The average value formula in the interval (a,b) is\n\nAV = 1/(b-a) * DEFINT f(x)  {evaluated from a to b}\nAV = 1/(b-a) * {INT f(4) - INT f(-4)} \n\nThe integral will be 1/2(x^4 - 5x^2).  If you evaluate this at x=4 and then at x=-4 and do the math it will be 0.\n\nSo the answer is 0.
830127
f''(x)=8x+4sin(x)\nf'(x)=4x^2-4cos(x)+c\nf'(0)=-4+c=4\nc=8\nf'(x)=4x^2-4cos(x)+8\nf(x)=(4/3)x^3-4sin(x)+8x+c1\nf(0)=c1=2\nf(x)=(4/3)x^3-4sin(x)+8x+2\n\nputting 5 into sin(x) is weird
210475
f(g(x)) = 3/( (3/x)+4) -4)\nso that becomes\n3/ ((3/x)) because the + 4 and the - 4 cancel out\nand when you divide you multiply by recipricol so 3 divided by 3/x = 3 multiplied by x/3\n3 multiplied (x/3) = 3x/3 = x
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;
321475
*(2x/(x^2-4)) + (5/2-x) - (1/2+x)\n*(2x/(x^2-4)) + (5/2-x)((2+x)/(2+x)) - (1/2+x)((2-x)/(2-x))\n*(2x/(x^2-4)) + (10+5x)/(4-x^2) - (2-x)/(4-x^2)\n*(2x/(x^2-4)) + (8+6x)/(4-x^2)\n\nyou should be able to get it from there
872722
F'(x) = √(5-x) - x/[2√(5-x)]\n\nF''(x) = -1/[√(5-x)] - 1/[2√(5-x)] + x/[4√(5-x)]³\n= - 3/[2√(5-x)] + x/[4√(5-x)³]
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
767821
P=F/A\nP=(4.31 N)/(5.07^10^-5 m^2)\nP=8.50x10^4 Pa
1005272
f(4) means 'rewrite f(x) = 2x - 5 with a "4" instead of an "x"', like this:\n\nf(4) = (2 * 4) - 5\n=8 - 5\n=3
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
1149923
(2,-4) if f(x) is defined for x=2
623250
substitute the x in the function for the 3.  that should give you:\n\nf(3) = (-5(3)^3)+(6(3)^2)-3-4  \n\nf(3) = (-135)+(54)-7   (simplified)        \n\nand your answer is:\nf(3) = -88
756998
f(g(x))= 4g(x) -5\n= 4(x/3) -5\n= 4x/3 -5\ns
551877
f(x) = sinx on [0,pi/2] is 1-1 on this interval, therefore there can be no mode, since all values occur only once.\n\nThe mean is equal to the area under f(x) = sinx on this inteval divided by the length of the interval.  In this case, it is 1/[pi/2-0] = 1/(pi/2) = 2/pi.\n\nThe mode is the value where half the area is below this value and half is above it.  Therefore, we want to know for what value of a is Integral from 0 to a of sinx = 1/2.  When you integrate and evaluate, you get cosa = 1/2, which means a = pi/3, so pi/3 is the median.
780025
x ≤ 5\n\n-- - - - - - -s-
360360
The average value of a function is given by\n\n(1/(b-a)) S f(x) dx, evaluated from a to b.\n\nSo for this question, the average will be given by\n\n(1/(4-0)) S 8+4e^-2t+te^-2t dt = (1/4)[8t - 2e^(-2t) + S te^-2t dt]\n\nLet u = e^(-t)\ndu =  -e^(-t) dt\n\nSo, S te^-2t dt = - S u^2 du\n= (-u^3)/3\n= -e^(-3t)/3\n\nSo your answer then becomes\n\n(1/4)[8t - 2e^(-2t) - e^(-3t)/3] = 2t - (1/2)e^(-2t) - e^(-3t)/12\n\nEvaluating from 0 to 4,\n= 2(4) - (1/2)e^(-2(4)) - e^(-3(4))/12 - [2(0) - (1/2)e^(-2(0)) - e^(-3(0))/12]\n= 8 - (1/2)e^(-8) - e^(-12)/12 + 1/2 + 1/12\n= 8 5/12 - (1/2)e^(-8) - e^(-12)/12\n~ $8.42
1102849
gurl am not helping out with ur homework next time.\n f(x) =/x/ +3x +1\n       =x+3x+1\n       = 4x+1\n\n  f(x)=3x+/2x-5/\n       = 3x+2x-5\n       =5x-5\n\n  g(x)= /x/+/x-1/\n        = x+x-1\n        =2x-1\n\n  g(x)=3/x-2/ - /x+1/\n        =3(x-2) - x+1\n        =3x-6-x+1\n        =2x-5