average value of f(x)=6x-x^2+16 on the interval [2,5] is?
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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/(5-2)) S 6x - x^2 + 16 dx = (1/3)(3x^2 - x^3/3 + 16x)\n\nEvaluating from 2 to 5,\n\n= (1/3)((3(5)^2 - (5)^3/3 + 16(5)) - (3(1)^2 - (1)^3/3 + 16(1)))\n= (1/3) 75 - 125/3 + 80 - 12 + 8/3 - 32)\n= 24\n\nAnswer D.
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gopal's solution:\n\nf(x)=x^3-6x^2+9x+2\nf'(x)=3x^2-12x+9=0 for extremas\n3(x^2-4x+3)=0\n3(x-3)(x-1)=0\nx=3 or 1\nf''(x)=6x-12\nat x=1 6(1)-12=-ve so max\nat x=3 6(3)-12=+ve so min\nso the absolute maximum is at x=1 and\n=6\nabsolute min is at x=3 and is=2  <------ However still need to check endpoints of the interval [-1,4] as these turnng points are actually LOCAL maxima and minima!\n\nSo f(-1) = -1 - 6 - 9 + 2 = -14 < 2\nSo ABSOLUTE minimum = -14 at x = -1\n\nAlso f(4) = 64 - 96 + 36 + 2\n= 6\nSo ABSOLUTE maximum = 6 and occurs at both x = 1 AND at x = 4
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Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
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1. f(x) = 200x - 15/40 - 2x\n\nI assume you mean\nf(x) = (200x - 15)/(40 - 2x)\n\n= 5(40x - 3)/[2(20 - x)]\n= 5/2 * (40x - 3)/(20 - x)\n= 5/2 * (40x - 800 + 797)/(20 - x)\n= 5/2 * -40(20 - x)/(20 - x) + \n= 100 + 5/2 * 797/(20 - x)\n\nNow we can deal with it, as messy as it looks as before it was 'improper' since the degree of the numerator was the same as the degree of the denomonator and a rational function is 'improper' if degree numerator ≥ degree denominator\n\nNow lets look at 5/2 * 797/(20 - x)\n\nA. When x < 20, 20 - x > 0 so 5/2 * 797/(20 - x) > 0. \nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) > 100\nAND when x > 20, 20 - x < 0 so 5/2 * 797/(20 - x) < 0. \nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) < 100\n\nie for x > 20, f(x) < 100 and for x < 20, f(x) > 100\n\nB. When x = 20 20 - x = 0 so 5/2 * 797/(20 - x) is undefined.\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) is undefined.\nAs x → 20 from the left ,5/2 * 797/(20 - x) → +∞\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ +∞\nAND as x → 20 from the right, 5/2 * 797/(20 - x) → -∞\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x))→ -∞\n\nie vertical asympotes at x = 20\n\nC As x → ±∞, 5/2 * 797/(20 - x) → 0\nTherefore f(x) = 100 + 5/2 * 797/(20 - x) (= (200x - 15)/(40 - 2x)) → 100\n\nie horitontal asymptotes at f(x) = 100\n\nD. You now have enough information to sketch this\n\n2. Work the function g(x)=x^2/x-2\n\nI assume you mean g(x)=x^2/(x - 2)\n\nAgain this is improper as deg numerator = 2 and deg denominator = 1 (< deg numerator)\n\nSo g(x) = x^2/(x - 2)\n= (x^2 - 4 + 4)/(x - 2)\n= ((x - 2)(x + 2) + 4)/(x - 2)\n= x + 2 + 4/(x - 2)\n\nDo steps A. to D as above\n\nA. Check out the signs of the denominator of the fraction\nWhen x < 2,  x - 2 < 0 so 4/(x - 2) < 0 \nTherefore g)x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} < x + 2\n\nWhen x > 2,  x - 2 > 0 so 4/(x - 2) > 0 \nTherefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)} > x + 2\n\nie for x < 2, g(x) < x + 2 and for x > 2, f(x) > x + 2\n\nB. When x = 2,  x - 2 = 0 so 4/(x - 2) is undefined.\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) is undefined.\nAs x → 2 from the left, 4/(x - 2) → -∞\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → -∞\nAND as x → 2 from the right, 4/(x - 2) → +∞\nTherefore g(x) = x^2/(x - 2) (= x + 2 + 4/(x - 2)) → +∞\n\nie vertical asymptotes at x = 2\n\nC. As x → ±∞, 4/(x - 2) → 0\nTherefore g(x) = x + 2 + 4/(x - 2) {= x^2/(x - 2)}  → x + 2\n\nTherefore g(x) is assymptotic to g(x) = x + 2\n\nD. You now have enough information to sketch this
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The answer is 10\nYou find the average rate of change of the function f(x) over a given interval, by computing the total change in the function f(x) over the interval, and then dividing this value by the size of the interval.\nYour interval size is 2- (-1) = 3. So you want:\n\n(f(2) - f(-1)) / 3  = (22-(-8))/3 = 30/3 =10\n\nf(2) = 2(2^3) - 2^2 +5*2 = 16-4+10 = 22\nf(-1) = 2(-1^3) - (-1^2) -5 = -2-1-5 = -8
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Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
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You need to graph this function. \nIt is a line.\ny = (x^2)^ 1/2 - 16 \ny = x - 16\nx = 0, y = -16, \nx = 16, y = 0
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1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1\nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n.
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ok.\n6/x-8 >= 5\n\n6 >= 5 (x-8)\n\n6 >= 5x-40\n6+40 >= 5x\n\n46 >=  5x\n\n46/5 >= x\n\nx <= 46/5
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I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
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x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
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x^2 - 8x + 16\n\n\nthe person above me said it is -x^2 but that is incorrect because -x * -x = x^2
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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.
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(a^2  - 5) * (a^2  - 2)
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You add the numbers together and then divide by the number of numbers...for example: 2, 6, 8, 4\n\n2+6+8+4 = 20\n20/4 = 5\n\nThe average of the 4 numbers is 5.
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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.
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f'(x)=-3+10x-3x^2\nf''(x)=10-6x\n\nmax and min are where f'(x)=0\n3x^2-10x+3=0\n(x-3)(3x-1)=0\nx=3,1/3\n\nto see if they are max or min use f''(x)\nf''(3)=-8   means concave down= max at x=3\nf''(1/3)=8   means concave up= min at x=1/3\n\ninflection points are where f''(x)=0\n10-6x=0\nx=5/3\nf''(0)=10 means concave up on the interval (-inf,5/3)\nf''(5)=-20  means concave down on the interval (5/3,inf)