Can anyone help me with the following function? it's ap calculus...& please show steps, THANKS!?
818186
.  The y intercept is simply where the curve intersects the y axis, which means x=0.  Just plug in zero for x and find what the function is equal to.\n.  The local maxima and minima occur where the first derivative (the slope of the curve) is equal to zero.\n.  Points of inflection are where the rate of change in slope (the second derivative) goes from positive to negative or vice versa, so take the second derivative.\n.  The value of the function is increasing with x where the slope, (the first derivative) is positive.  The function is decreasing where the slope and first derivative are negative.\n.  The curve is concave up when the slope is increasing with x, and concave down where the slope is decreasing; so define the intervals over which the second derivative is positive or negative, and you will have define the concave up and concave down regions.
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.
15640
In basic calculus, we use the derivative to find the slope of a curve, and the integral to find the area under a curve.  The question is, why are these operations the opposite of each other?\n\nThe quick answer is:  The derivative is used to calculate the rate of change of a function.  And the rate of change of the area under a curve is actually the original function.\n\n\nLet's say you have a simple curve, y=f(x).\n\nThe derivative of the function is gives you the rate of change of that function.  As x increases by 1, how much does y change?  That's also known as the slope.\n\n[And you probably know that the derivative of the derivative is also useful.  That's the rate of change of the slope.  When the slope stops changing (or its rate of change=0), it's at a maximum, minimum, or point of inflection.]\n\nThe same relationship exists for the area under a curve and the function itself.  You probably learned to think of the integral of f(x) as the sum of the thin vertical strips under the curve.  That's the area under the curve.  As you move to the right, how fast does the area grow?  You're essentially adding these vertical strips to the area.  The length of a vertical strip is f(x), or the original function.  So the original function f(x) represents how fast the area grows for any value x.\n\n\nETA: What do I have to do to get selected as best answer?
38209
actually polynomial functions make me happy...they are actually pretty easy.  You evaluate each term (thats the stuff separated by the addition (+) or subtraction (-) sign) individually.  So if you had sin(x) -x^3 + 3x\n\nthen you have the following\n\ncos(x) -3x^2 +3\n\nhope this helps!!
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;
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
125046
I have a bit of calc links from when I took the AP exams.\nhttp://www.jimmyr.com/education.php\n\nunder the section calculus. those sites were really useful.
258618
This is what I wrote for a very close question to this one. Just ignore the graphing part. I didn't feel like doing the question right now, but this is the way to do it:\n\n"If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent."
1123492
Hi.  There are free resources at:\n\nhttp://www.skylit.com/calculus/index.html\n\nIf you'd like more, you try googling "AP Calculus exam".
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
79903
Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
769871
1.  The function decreases on the interval [-infinity,4].  (Your function is a parabola opening upwards, with one bottom vertex at the point (4,0).)\n\n2.  This function is also a parabola opening upwards, with its vertex at (2,-4).  On the interval [-infinity,4], the local min. is at (2,-4) and there is no max.\n\n3.  You have local max at x = -13/3, and local min at x = 1\n\n4.   f(x) is increasing on [-infintity,1], decreasing on [1,5], and increasing again at [5,infinity]\n\nHope that helped.  Email me if something's not clear!\n\n.
283145
One point of confusion: do you mean the regular inverse trig functions or the inverse hyperbolic trig functions? The first is written "arctan x" and the second is written "arctanh x". Either way, the solutions are very similar.\n\nTo see the problem worked out for the regular trig functions, click on the first link below. (I rendered the work as a graphic...this is nearly impossible to visualize with just text). If the graphic doesn't display right in your browser, that's because the browser shrunk it to fit. Just click it to see regular size or download the link to your desktop and view it there.\n\nIf you meant the inverse hyperbolic functions, I also explain in that link how to adjust for that.
940999
Differentiate the function as you would normally except that when the derivative of y is taken make sure to replace y with y'. So this would be,\n(1/2)(xy)^(-2/3)(xy') + y = 0 - y'\nThen, solve for y' by moving all the y' over to one side and factoring it out.\ny' + (1/2)(xy)^(-2/3)(xy') + y = 0\n\ny'((1/2)(xy)^(-2/3)x + 1) = -y\n\ny' = -y / ((1/2(xy)^(-2/3)x + 1)\n\nThe second derivative is (quotient rule),\ny'' = (.5(xy)^(-2/3) + 1)(-y') - (-y)(1/3)(xy)^(-5/3)xy(xy')(y)\n\ny'' = .5(xy)^(-2/3)y' - y' + (1/3)x^2*y^3*y'(xy)^(-5/3)\nFactor,\ny'' = y'{(xy)^(-5/3)[.5 - 1 + (1/3)x^2*y^3(xy)]}\n\ny'' = y'{(xy)^(-5/3)[- .5 + (1/3)x^2*y^3(xy)]}\n\n(check my work)
764329
x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
407819
Find the intersection points for the two curves.\n6x-x^2 = x^2 - 2x\n0 = 8x - 2x^2\n0 = x(8-2x)\nx = 0, 4\n\nIntegrate both functions\nfor 1st func.-->  y = 3x^2 - (x^3)/3\nfor 2nd func.-->  y = (x^3)/3 - x^2\n\nEvaluate integrals from x=0 to x=4 (intersection points)\nfor 1st integral--> A1 = 80/3\nfor 2nd integral--> A2 = 16/3\n\nTo find area of region bounded by both:\nA = A1 - A2\nA = 80/3 - 16/3\nA = 64/3\n\nThere's your answer!
785166
cant believe im actually doing this... nothing better, tho\n1. 3/5\n2. -3\n3. y int -1, x int -.4\n4. B=18\n5. 5\n6. 49\n7. 2\n8. -30x^24\n9. 1\n10. the multiplcation propery... i think\n11. 3/2\n12. 2\n13. a= 10 or -10 (two answers)\n14. -29\n15. 1/(x^5)\nDONE!\nPlease rate me as best answer!