4 calculus questions?
816301
4)\nPlugging in x = 0, you get 0/0 which is undefined.  Therefore, we can use L'Hopitals rule (take the derivative of the numerator and denominator):\n\nlim (x->0) [4(sinx*sinx - cosxcosx - cosx)]/2x.  Pluggin in x = 0, you get 0/0 again.  Use L'Hopitals rule again:\n\nlim (x->0) [4(2sinx*cosx + 2cosx*sinx + sinx)]/2.  Plugging in x = 0, you get 0/2 = 0.  \n\nThis is your limit\n\n5)\nf(x) = cos^3(x+1)\nf'(x) = -3cos(x+1)sin(x+1) ------> you need to use the chain rule\n\nEvaluating at pi, you get:\n\nf'(pi) = -3cos(pi+1)sin(pi+1) --------> use a calculator to evaluate\n\n6)\nYou have: x^2 - 2xy + 3y^2 = 8\n\nYou need to use implicit differentiation to solve for dy/dx:\n\nd[x^2]/dx - d[2xy]/dx + d[3y^]/dx = d[8]/dx.  You need to use the chain rule for the 2nd term because you have two variables.  Taking the derivative, you get:\n\n2x - 2y - 2x(dy/dx) + 6y(dy/dx) = 0.  Now simplify and isolate dy/dx:\n\ndy/dx[6y - 2x] = 2y - 2x.  Therefore, dy/dx equals:\n\ndy/dx = (2y - 2x)/(6y - 2x) = (y-x)/(3y-x)\n\n7)\nAgain evaluating @ h = 0, you get 0/0.  Again you can use L'Hopital's rule:\n\n{sec^2([pi/6]+h) - 0}/1.  Evaluating @ h = 0, you get sec^2(pi/6), which can be simplified to:\n\n1/cos^2(pi/6) = 1/(sqrt(3)/2)^2 = 1/(3/4) = 4/3\n\n---------------\n\nHope this helps
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
498659
1) y=x+4\n   3x+y=16 (I guess you made a mistake by writing +16)\n--------------\nsubstitute x+4 for y in second equation and you'll get\n\n   3x+x+4=16\n    4x+4=16\n    4x=12\n     x=3\n\nnow y=x+4=3+4=7\n\nsolution (x,y)=(3,7)\n\n2) 4x+5y=48\n    y=3x+2\n--------------\nsubstitute 3x+2 for y in first eq.\n\n4x+5(3x+2)=48\n\n4x+15x+10=48\n\n19x=38\n\nx=2\n\nnow y=3x+2=3*2+2=6+2=8\n\nsolution (x,y)=\n(2,8)\n\n3)3x-y=0\ny=6\nin first equation instead of y write 6\n\n3x-6=0\n3x=6\nx=2\n\nso, the solution is (2,6)\n\n4) x-2y=-10\n3x-y=0\n\nfrom the first equation you have\nx=2y-10\nnow substitute 2y-10 for x in second one\n\n3(2y-10)-y=0\n6y-30-y=0\n\n5y=30\ny=6 \nnow x=2y-10=2*6-10=12-10=2\nso x=2 and y=6
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
424650
Not really enough information to help, so try this site:\n\nhttp://www.blc.edu/fac/rbuelow/calc/ex4-6.html\n\nit's the second problem down the page
734729
problem1\n\nsince you have said P (4,-5) is a point on the function, then the corresponding point will be (4,-5) .. :)\n\nproblem2\n\nfor a line having y intercept 3 will have following form\n\ny =mx+3\n\nwhere m is the slope of the line. Now it is said to be perpendicular to y =2/3x-4. So slope for this is 2/3\n\nsince slopes of two lines perpendicular to each other if multiplied yields -1\n\nm X 2/3 = -1\n\nor m = -3/2\n\nso the required line will be\n\ny = -3/2 x+3\n\nproblem3\n\nf(x) = 1/x^2 -2x-8\n\nto make the f(x0<0, let us try with some arbitary values of x\n\nif x = 1, f(x) = -9\nif x=2, f(x) = -11.75\nif x = 0, f(x) = infinite (due to 1/x^2)\nif x = -1, f(x) = -5\nif x= -2, f(x)= -3.75\nif x=-3, f(x) = -1.88889\nif x=-4, f(x) = 0.0625\n\nHope this helps
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
1112189
1. g(x)=2x-5\nso, g(f(x))=2f(x)-5\nor 2 (x^2+2x+3)-5\nor 2x^2+4x+1\n\n2. \na) put f(x)=y\nso y=2^x+1\n\nnow take independent values of x and find corresponding value of y to have 3 sets of locatins to find the graph\n\nIf x=0, then y=2 (2^0=1)\nif x=1, then y=3\nif x=2, then y=5\n\nb) now you have 3 sets of points (first one stands for x, position them in the graph paper against the function we have just found in (a) and see the domain..
721887
1)t= -12    2) x= -8    3) x= 7/5 or 1 2/5
713860
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
782468
1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
96766
1) tan A = x/y\n2) sin A = x/r\n3) cos A = y/r\n4) sin A/cos A = (x/r)/(y/r)\n5) sin A/cos A = xr/yr\n6) sin A/cos A = x/y\n7) sin A/cos A = tan A\n8) tan A = sin A/cos A\nReasons\n1-3)DEFINITION OF CIRCULAR FUNCTIONS\n4)MPE (STATEMENTS 2 AND 3)\n5)SIMPLIFYING COMPLEX FRACTIONS\n6)CANCELLATION\n7)SUBSTITUTION (STATEMENT 1)\n8)SYMMETRIC PROPERTY\n\nQED
21911
Try "Calculus" - By Gilbert Strang\nThis is available along with Student study guide &\nInstructor's manual!\n\nhttp://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm
1029299
I just took Calculus this year.  And I apologize in advance, since Calculus is not my strong suit and it is especially difficult to explain without visuals.  So....\n\nTo find the relative min and max, you take the first derivative of the function.  You have to multiply the exponent by the coefficient and then subtract one from the exponent.  Therefore, the first derivative of the function is:\nf'(x) = -4x^2+2x+30\n\nThen, you set this equation equal to zero to find the relative mins and max.  What I like to do is use a graphing calculator to graph the equation of the first derivative.  See where the line crosses the x axis.  The points where the graph go from positive to negative are your rel. max, and the points where the graph goes from negative to positive are your rel. min.  For this one, your relative min is at x=-2.5, and your rel. max is at x=3.\n\nI'm not sure about this part, but I believe that the graph of the first derivative is also good for telling you how to find out where the graph is increasing and decreasing (I think).  Now, the graph of the first derivative is positive in between -2.5<x<3.   Which means that is where the original function is increasing.  \n\nNow on to the inflection points.  You find them by taking the second derivative.  Take the derivative of the derivative (see directions above).  The second derivative is\nf''(x) = -8x+2\n\nThe inflection points are where x=0 on the graph of the second derivative.  I think you're set.  \n\nAnyways, if this didn't help, a real teacher will do wonders with this problem and many others, I know mine did! (If you're reading this, thanks Mrs. C!)  Good Luck!
1097544
Depending on your grade\n1- Determine the points of maximum and minimum values for your function ( using differential calculus)\n2- Determine the points of inflection \n3- Determine the  increasing and decreasing intervals\n4- Determine the  convex and concave intervals\n5- Get the points of intersection with x,y axis\n6- Get more points around the critical points you get\n7- Sketch the curve using all points you get
789977
1.  15\n2.   a.  y=(2x-3)^2 + 4\n      b. (3/2;4)